DIEN DAN TOAN HOC VIET NAM www.maths.vn OQo Tuyền tập Bat đăng thức Volume 1 Biên tập: Võ Quốc Bá Cần
Tác giả các bài toán: Trần Quốc Luật Thành viên tham gia giải bài:
1 Võ Quốc Bá Cần (nothing) 2 Ngo Dire Loc (Honey_ suck)
3 Trần Quốc Anh (nhocnhoc)
4 Seasky
Trang 2Tran Quoc Luat’s Inequalities
Vo Quoc Ba Can - Pham Thi Hang
Trang 3i
Trang 4Preface
"Life is good for only two things, discovering mathematics and teaching mathematics.”
S Poisson
Bat dang thuc la mot trong linh vuc hay va kho Hien nay, co kha nhieu nguoi quan tam den no boi no thuc su rat don gian, quyen ru va ban khong can phai "hoc vet" nhieu dinh ly de co the giai duoc chung Khi hoc bat dang thuc, hai dieu cuon hut chung ta nhat chinh la sang tao va giai bat dang thuc Nham muc dich kich thich su sang tao cua hoc sinh sinh vien nuoc nha, dien dan mathsvn da co mot so topic sang tao bat dang thuc danh rieng cho cac ca nhan tren dien dan Tuy nhien, cac topic do con roi rac nen ta can mot su tong hop lai thong nhat hon de cho ban doc tien theo doi, do la li do ra doi cua quyen sach nay Quyen sach duoc trinh bay trong phan chinh bang tieng Anh voi muc dich giup chung ta ren luyen them ngoai ngu va co the gioi thieu no den cac ban trong va ngoai nuoc Mac du da co gang bien soan nhung sai sot la dieu khong the tranh kho1, rat mong nhan duoc su gop y cua ban doc gan xa Moi su dong gop y kien xin duoc gui ve tac gia theo: babylearnmath@yahoo.com Xin chan than cam on!
Quyen sach nay duoc thuc hien vi much dich giao duc, moi viec mua ban trao doi thuong mai tren quyen sach nay deu bi cam neu nhu khong co su cho phep cua tac gia
Vo Quoc Ba Can
Trang 6Chapter 1 Problems "Each problem that I solved became a rule, which served afterwards to solve other problems." R Descartes 1 Given a triangle ABC with the perimeter is 2p Prove that the following inequality holds a b é b+c cta a+b +——Ừ+ >i TC + tr (EE,
p-a p-b p-c p-a p—b p-e
2 Let a,b,c be nonnegative real numbers such that a? + 5? + c? + abe = 4 Prove that the following inequality holds a+h+ec > a?b? + bˆc? + cđ 3 Show that for any positive real numbers a,b,c, we have a+bh+c+b6abe > Vabc(a+b+e)
4 Let a,b,c be nonnegative real numbers with sum 1 Determine the maximum and minimum values of
P(a,b,c) =(1+ab)? + (1+ hc)? + (1+ ca)?
5 Let a,b,c be nonnegative real numbers with sum | Determine the maximum and minimum values of
P(a,b,c) = (1 —4ab)* + (1 — 4c)? + (1 —4ca)’
6 Let a,b,c be positive real numbers Prove that
b+c eta atb ? 1 1 1
Trang 710 11 12 13 14 15 16 Problems Let a,b,c be the side of a triangle Show that 3 (a+Ð)(a+e)Vb+e—a>4(a+b+e)v(b+ec—a)(c+a—b)(a+b— c) cyc Given a triangle with sides a, b,c satisfying a* + 5? + c* = 3 Show that at+b b+c c+a >6 Va+b-c VWb+c-a Vvc+a-b_ ` Given a triangle with sides a, b, c satisfying a” + b? +c? = 3 Show that b ——+ +——— 33 Vb+c-a vec+a-b va+b-c Show that if a, b,c are positive real numbers, then a + b + c S14 2abe
a+b b+c cta™~ (a+b)(b+c)(c+a)
Show that if a, b,c are positive real numbers, then
a " b " c "3, Pb+ bet Ca—3abe
a+b b+c c+a/j — 4 (a+b)(b+c)(c+a) `
Let a,b,c be positive real numbers Prove that
(+b \(b +e) +a’) e+e +2 \7 §a2b7c2 — \ab+bc+caj/ `
Let a,b,c be positive real numbers Prove the inequality
(b+e)" (c-+a)" (a+)
a(b+c+2a) b(c+a+2b) c(a+b+2c) —
Let a,b,c be positive real numbers Prove the inequality
(b+e}? (eta) (a+b)? ` b+c 1 c+a 1 a+b
a(b+c+2a) b(c+a+2b) c(a+b+2c) — \b+c+2a c+a+2b a+†b+2cj,
Let a,b,c be positive real numbers Prove that
(b+c) (eta) (a+b)? sof 4 b "_
a(b+c+2a) b(c+a+2b) c(a+b+2c) — \b+c ct+a a+bj
Let a,b,c be positive real numbers Prove that
Trang 817 18 19 20 21 22 23 24 If a,b,c are positive real numbers such that abc = 1, show that we have the following inequality 3.43, 3 a b C 3
a+b`+ể>——+——+_——+ b+c cta atb 2
Given nonnegative real numbers a,b,c such that ab+ bc+ca+abc = 4 Prove that
a +b? +c? +4+2(a+b+c)+3abe > 4(ab+be + ca)
Let a,b,c be real numbers with min {a,b,c} > 7 and ab + bc + ca = 3 Prove that
atbh+c+9abe > 12
Let a,b,c be positive real numbers such that a*b* + b*c? +¢7a? = 1 Prove that
(a? +b? +0°)? +abey/(a2+b2+02)3 > 4
Show that if a,b,c are positive real numbers, the following inequality holds
(a+b+ 6) (ab + be + ca)Ÿ + (ab + be + ca)3 > 4abe(a-+b+ e)Ÿ
Let a,b,c be real numbers from the interval [3,4] Prove that
b b
(atbt+e) (PO 4 ~4")\ 53/240? +2), C a b
Given ABC is a triangle Prove that
8 cos” A cos’ Bcos”C + cos2A cos2Bcos2C > 0
Let a, b,c be positive real numbers such that a+ 6+ =3 and ab+ be+ca > 2 max {ab, bc, ca} Prove that
Trang 10Chapter 2
Solutions
"Don’t just read it; fight it! Ask your own questions, look for your own examples, discover your own proofs Is the hypothesis necessary? Is the converse true? What happens in the classical special case? What about the degenerate cases? Where does the proof use the hypothesis?”
P Halmos, / Want to be a Mathematician
Problem 2.1 Given a triangle ABC with the perimeter is 2p Prove that the following inequality holds a + —— pte a a+b p-a p- "5+ pc ¬ at ‘Vp ~c Solution Setting x = p—a,y= p—bandz= p—c, thna=y+z,b =z+x andc=x+y The original inequality becomes +Z Z4+X xX+ +Z Z+X x+
yee poy s/o peg op x y Zz x y fog Zz
Trang 116 Solutions
or equivalently,
y+Z + z†# + x1 > 6, x y Zz which 1s obviously true by AM-GM Inequality
Equality holds if and only ifa=b=c L] Problem 2.2 Let a,b,c be nonnegative real numbers such that a* + b? +c? + abe = 4 Prove that the fol- lowing inequality holds a?-+-bˆ+c > eh +h +a’ Solution From the given condition, we see that there exist x,y,z > 0 such that 2x b 2y 2z ——————, —= ———, €—_——— (x+z)(x+z) +z)+*) (z+x)(z+y) Using this substitution, we may write our inequality as 4y?z?
Leer ary = Lore ee
which 1s obviously true because
yˆzZ? yz(y+z)Ÿ
>przeinet SJ6r2eler3 = VEDjer3” Cục cyc Cục
and
Ler ary Leper
Our proof is completed Equality holds if and only if a=b=c=1 ora=b= vV2,c =0 and its cyclic permutations LÌ Problem 2.3 Show that for any positive real numbers a,b,c, we have a+h+c+46abe> Vabcla+b+c) Solution I, According to the AM-GM Inequality, we have the following estimation 6Ÿ⁄abe(a+b+e)? < (a+b+e))+-9Ÿ42Ð2c?1(a+ b+ )
which leads us to prove the sharper inequality
6( +b> +03 +6abc) > (at+b+cP4+9VeR (a+b +c), or
S(a +b> +07) — 3) ab(a+b) + 30abe > 9Ÿ222c2(a+ b+ e),
cyc
From Schur’s Inequality for third degree, we have
3(a +b? +03) — 33 _ab(a+b) + 9abe > 0,
Trang 12and we deduce our inequality to
2(a° +b? +0?) +2labe > WVeR(atb+e),
Again, the Schur’s Inequality for third degree shows that
4(a` + bŸŠ+c})+ 15abe > (a+b+e)Ÿ,
and with this inequality, we finally come up with
(a+b+c)? — Sabe +42abe > 18V.a2b2c2 (a+b +c), (at+b+c)> +27abe > 1842b2c2(a-+b+ c),
By AM-GM Inequality, we have that
2(a+b+c)> + 54abe (at+b+c)> + [(at+b+c)* +27abe + 27abc| (a+b+e))+27a2»2c2(a+b¬+c) 9Ÿ/222c2(a+b-e)-+27Ÿa2b2c2(a+b+ e) 36VERC(a+b+e) IV IV It shows that
(at+b+c)> +27abe > 1842b2c2(a-+b+ c),
Trang 138 Solutions
So, the above statement holds and all we have to do is to prove that P(t,t,c) > 0, which is equivalent to each
of the following inequalities 27 +46 > (2t+c)?, 3 1 1\? 2° +—+6> (2145) , t t 2/7 +6/° +1>P (27 +1), 2/2+6/®+ 1> 4Š +4 +12, 2/2 — 4fŠ-+6/®—4i—r?+1>0, — Ÿ (217 28 424 +28 42° 4204 1) >0,
which is obivously true because / > 1 L] Problem 2.4 Let a,b,c be nonnegative real numbers with sum | Determine the maximum and minimum values of
P(a,b,c) =(1+ab)? + (1+ bc)? + (1+ ca)’
Solution It is clear that minP = 3 with equality attains when a = 1,b = c = 0 and its cyclic permutions Now, let us find max P We claim that max P = vi attains when đ — 5 = c=— 3 OF 100 (1 +ab)? + (1 +bce)? + (1+ ca)? < Te 19 aˆbŸ + bˆcˆ + c°a? + 2(ab + be + ca) < 57° 3 19 (ab + bc + ca)“ + 2(ab + be + ca) — 2abc < 7 2 46
(ab + be + ca-+ L)“ — 2abec < 7
According to AM-GM Inequality, we have (ab+be+ca+1)* < =(ab+be+ca+1) wo] And we deduce the inequality to 4 46 3 (ab + be +ca + 1) — 2abc < 7 10
A(ab + bc + ca) — 6abc < 9°
which is obviously true by Schur’s Inequality for third degree,
1 10 A(ab + bc + ca) — 6abe < (1+ 9abc) — 6abe = 1+ 3abe < 1+3: WO
Trang 14Problem 2.5 Let a,b,c be nonnegative real numbers with sum | Determine the maximum and minimum values of
P(a,b,c) = (1 —4ab)* + (1 —4be)? + (1 —4ca)?
Solution (by Honey_suck) Notice that 1>1-—4ab>1-—(a+b) > 1-(a+b+c) =0 Hence (1 —4ab)? <1, and it follows that P(a,b,c) <1+1+1=3,
which equality holds when a = 1,b = c = 0 and tts cyclic permutions And we conclude that max P = 3 Moreover,applying Cauchy Schwarz Inequality and AM-GM Inequality, we have 1 (1 —4ab)? + (1 —4bc)? +(1—4ca)?_ > 3 (1 —4ab +1 —4be+ 1 —4ca)” 1 = 33 —4(ab + be + ca)’ 1 1\? 25 > =-(3-4.-) == — ;Í ;) 27
with equality holds when a=b=c= 3 And we conclude that min P = 3
This completes the proof LÌ
Problem 2.6 Let a,b,c be positive real numbers Prove that b+ + +b ( ¬ cle a 5 + a ) >4(eð+be+ea) (TS +555) : 1 11 bˆ_ c? Solution I, We can rewite the inequality as 2 > A(ab+be+ca)(a?b*? +b’? +a’) Eon (a+b) cọc
Assuming that a > b > c, then using AM-GM Inequality, we have that
A4(ab + be + ca)(aˆbŸ + bˆc? + c?a”) = 16(a+b)?(ab+ be+ca)(a?b* + bc? + ca’) A(a+b) [(a+ 6)? (ab + be + ca) +4(a°b? + bc? +.c7a’)| ° — A(a+b)
It suffices to show that
Trang 1510 Solutions 2 J,2 2 2 2 2 2ab(a+ b) +2c*(a+b) + 2c(a’ +b”) > (a-+6)(ab-+be-+ ca) +A Fe Fee) a 2 p2 2 (22 L p2 ab(a+b) +2c?(a+b)+c(a—b)* > 4a Aca +6") — a+b a+b_ ` 4ab 2(a? +b? ab(at+b—-— +c(a— b) > 2c? a+) oy ; a+b a+b ab(a—b)? 5 2c?(a—b/ pe eee _pr>TY a+b tela— by 2 a+b ` ab te> 2c? a+b ` “a+b` which 1s obviously true because 2c? 2c? ab < =c<ct+
a+b~ ct+e a+b
The proof is completed and we have equality iffa =b=c L] Solution 2 Similar to solution 1, we need to prove 2 > A(ab+be+ca)(a?b*? +b’? +a’) 3 ab(a+b) cục
For all real numbers m,n, p,x, y,z, we have the following interesing identity of Lagrange
(x+y +27) (me? +? + p) — (mx+ ny + pz)” = (my — nx)’ + (nz — py)? + (px— mz)’
Trang 16II
Yah (a—b)* + 2abcÀ_a(a— b)(a— e) > 0,
cyc Cục
which is obviously true by Schur’s Inequality for third degree L] Problem 2.7 Lef a,b,c be the side oƒa triangle Show that 3 (a+Ð)(a+e)Vb+e—a>4(a+b+c)v(b+e—a)(c+a—b)(a+b— c) Cục Solution The inequality can be rewritten in the form b) y (a+ b)(a te) >4(a+b+e), œ/(e+a—b)(a+b—
which 1s obviously true because
(a+b)(a+c) = yetoyvare) (a+ b)(a+c) (cta—b)(a+b—c) oe a? — (bc)? y erate) 2 cục a be ca ab = A(atø+o)+ (+ +) ab c > 4(a+b+c), where the last inequality is valid because be | ca ab _ pe +6(£4+2)+ a? abc “\2e" 3b 2a 2c) '°\ 2b" 2a > atb+te
Equality holds iffa =b=c L]
Problem 2.8 Given a triangle with sides a,b,c satisfying a” + bˆ -} cˆ = 3 Show that
a+b b+e c+a >6
Va+b-c Vb+c-a Vvc+a-b_ `
Solution Firstly, to prove the original inequality, we will show that!
4a(b+e—a) 4b (cta—b) 4c(at+b—c)
(b+e) (c+a)’ (a+b) 7 ab+be+ca> 4a’ (b+ce—a) |e ere Cục >a? +b? +? —ab—be—ca,
'We may prove this statement easily by using tangent line technique, the readers can try it! In here, we present a nonstandard proof
Trang 1712 Solutions a’(2a—b—e)* | bY(2b~e~ a)’ c2(2e— a— b)? (0+) (c+a) (a+5)? Without loss of generality, we may assume that a > b > c, then >a +b +0? —ab—be-ca, a ` b2 (b+e}ˆ — (e+a)? Thus, using Chebyshev”s Inequality, we have and (2z—b—e)?>(2b—e—a)Ẻ a’ (2a—b—c)* b*(2b—e-a)? _ 1 a ?? 2 Do T— Gia 2|01271(eta|lŒs=ð=o+(s=c~=z)] Œ) Notice that 5 [(2a—b 0)? + (26 ~a)"] ~ (a? ab + 8°) > (a+b ~20)? ~ F(a +b) (2) And a ?ˆ ` 2(a+b)* =
(b4+e? | (chap ~ (a+b420) ~ 2’
which yieds that 1 : b° 1 3 rae ray 3] [Oe b-eP +0-e- a] 2 I[ 2(a+by 1 > ; [(2a—b—c)* + (2b -c—a)’] I[ 2(a+b) 1 > Ít rao? 2|(e+19~29Ẻ (3)
From (1), (2) and (3), we obtain
a’(2a—b—c) _ b(2b-c-a)? —(a’ —ab+b’) > (a+b)*(a+b— 2e)?
— x(a+B}
(+e — (c+a? a+b 420)
Using this inequality, we have to prove (a+Ð)7(a+b—2e) 1 2 €(a+b—2c) _ 5 ae ae 3a+p+2o5 at) +—~ Gap SC (+0), (ạibp21 Ác 5 2 b 2 _ 222 ¿2 —2¿\2 (a+b)“(a+b—2c)“ _ đ(a+b—2c) > Tq„+p—2e) 2(a+b+2c) (a+b) 4 2 2 (a+b) "- > I 2(a+b+2c)? (a+b)? ~ 4’
which can be easily checked Thus, the above statement is proved Now, turning back to our problem, using Holder Inequality, we have
2
b+c a(bte
Trang 1813 It follows that œvb+c-=a # (be 4) (Brey? ` ác Lab 3 3 4 In b+e ) 25 (ze) (E2) X
Moreover, by AM-GM Inequality, we have
Ya = 3 (Le) (Le) (De) 3 Yab+Yab+ya\? (x2) < 1 cyc cyc cyc _ AỚc — \ 3 3 9 Hence 3 2 4 ( » a) „_——— b +€ > aes cyc > 36 cyc V b +€C— — Lab cyc »
Our proof is completed Equality holds if and only ifa@=b=c=1 L] Problem 2.9 Given a triangle with sides a,b,c satisfying a’ + b? +c? = 3 Show that a b C + + > 3 VWb+c-a vwc+a-b va+b-c - Solution (by Materazzi) Applying Holder Inequality, we obtain 2 bai] lEeee-|eeee And we deduce our inequality to show that (at+b+c)> 9)" a(b+c—a), cục
(a+b+c) > 9[2(ab + be+ ca) — 3)
Trang 1914 Solutions Problem 2.10 Show that ifa,b,c are positive real numbers, then a + b + c S14 2abe a+b b+c c+ta™~ (a+b)(b+c)(c+a) Solution The given inequality is equivalent to 2abe 2abe T2 at B+o(e+a) Ì a+901+o(+a) Pu aha 2) = cyc cyc Using the known inequality? 2abe Yop _ (a+b)(b+e)(e+a)' cyc we can deduce it to
y— ` 2abe + 2abc
Gab) are? (a+b)(b+c)(c+a) (a+b)(b+c)(c+a)’
Cục
abh+b’ce+ca+abe > \/2abc(a+b)(b+e)(c +a) Now, we assume that c = min {a,b,c}, applying AM-GM Inequality, we have ad °b+b°c+ca+abec = alab+c*)+hc(a+b) +c)(b+ — a(a + ©)(b+‡ ©) x 9) + belatb) + a(a+c)(b+c) 2 > 2M IOS) pola pn = /2abc(a+ b)(b+c)(e+a) a(aT— c)(b — c) 2 > + bc(a+b)
Trang 2015 Hence, the given inequality can be rewritten as
Tụ 4abc ` ab + be + ca
4 (a+bð)(b+c)(c+a) —(a+Ð)} (b+c) (c+a)?
a—b\' + b—c\? +(——} >2-— _ c—a\ l6abe
a+b b+c c+a (a+ b)(b+c)(c+a) l6abe _ (a+b)(b+c)(c+a) 2|[(a+ b)(b+c)(c +a) — 8abc| (a+b)(b+c)(c+a) 2l + c)(a— b)(a— e) + (c+a)(b— e)(b —a) + (a+ b)(e— a)(e— 4) (a+ b)(b+c)(c+a) _ 2 | aes (b—c)(b—a) oe):
(at+b)(at+c) (b+c)(b+a) (c+a)(c+bÐ) Now, notice that The above inequality is equivalent to (1 0) Ea a—-b b—c ec—a\? + + = 0,
ath b+e cta
which is obviously true Equality holds if and only if@ = 6 or b=c orc=a L]
Problem 2.12 Leta,b,c be positive real numbers Prove that
(a +b*)(b? Fe*)(c* +a") 5 C4+RP4+e2\" §a2b2c2 — \ab+be+caj/ ` Solution I Without loss of generality, we may assume that a > b > c, then we have that 212 — ,\2/q2_ /”2_ 2_—_ + a +c )— |a + ——— = 2 0, d2 + B2\(g2 + c2 2 ore (b—c)* (8a ~ c* — 6bc) — 2 ` It suffices to prove that [4a? + (b+c)?] (b+c) ` a?+bˆ+c? l6abe — ab+ bc-+ ca
This inequality is homogeneous, we may assume that b+ c = 1, putting x = bc, then a > 3: and ; >x>0 The above inequality becomes
4a* +1 ` a +1—2x
l6ax ~ a+x
Trang 2116 Solutions
(4a? + 1)(a+x) > 16ax(aˆ + 1— 2x), 32ax? — (16a` — 4a” + 16a— 1)x+a(4aˆ +1) >0,
2a(4x— 1)? + 2a(§x— 1) — (16a` — 4a” + 16a— 1)x+-a(4a” +1) >0,
2a(4x — 1)? + (1+ 4a’ — 16a*)x + a(4a? — 1) > 0, 2a(4x — 1)? — (2a— 1)(8a” + 2a + 1)x+a(4a? — 1) > 0,
which 1s true because
—4(2a —1)(8a* +2a4+ 1)x+4a(4a?—1) > 4a(4a* —1)— (2a—1)(8a* +2a+4+1) (2a— 1) > 0
Our proof is completed Equality holds if and only 1Ÿ a = 5 = e L] Solufion 2 Put a — x,Ð = pre = 2 then the above inequality becomes
(x+y+2)“(ˆ°+y)0ˆ +2) +37) > 80 +r 42x)
Notice that
(WP 4VIVP F224) =P FV 4ZICV FVLP 47 P)— Ve
Thus, we can rewrite the above inequality as
(eV tyr +27x°) [x+ty +2)“ +y +2”) 822 ty 22 +z+”)| >x43//2ˆ(x+ey +2)”
Now, we see that
(x+ty+2) “+ /+z7) —8(x 9” +y2Z +z7x”)
= yx" + 2) xy(x7 +y")+ 2xyz) x — 63x” y
cyc cyc cyc cyc
= yx (x—y)(x—z) +3) xy(x — yy? +xyz) x > xyz) x
cyc cyc cyc cyc
It suffices to prove that
xyo(xry $y Ze +2x°)(x+by+z) > xy (xtytz/’,
Py +y22 4+27x? > xyz(xty+z),
which 1s obviously true by AM-GM Inequality L] Solution 3 Similar to solution 2, we need to prove that
(FYI +22 42°) (ety tz) > 80P¥ tye +2")
By AM-GM Inequality, we have
(x+y+z) = x7 4y? +27 42xy+2yz+ 22x
4x7? 4y7z? 4zˆx?
Pye pap 21x
Trang 2217 Hence, it suffices to prove that 4mn 4np cn 2 m+n)(n+ +m)|m+n+pt+ + + > 8(mn+np+ pm)’, (m-en\(n-t py(p-+m) (mens p+ 0 4 SIP 4 TT > 8m +np + pm) 2 2 where m=? ,n=b?,p=c’ This inequality is equivalent with mn(m—n)* +np(n— py + pm(p—m) > 0,
which is obviously true LÌ
Solution 4 ” Again, we wIll give the solution to the inequality
(FYI +22 42°) (ety tz) > 80P¥ tye +2")
Assuming that x > y > z, then by Cauchy Schwarz Inequality, we have @ +z02+Z) > (w+2) Hence, it suffices to prove that x2+y2)\(xy+z2)?(x+y+z)? 3 y > §8(x2y2-+y2z2 + z2x2), V20P + )avt2)(ety+z) >4G°y +2" +2°x°), We have x—y) (x—y)? V20? +) = xty+¢—— Oo x+y+ V2 +3”) sey yy —_ x+y+V2(x+y) 2} — (v2-1)—3Ÿ 3 —y} = TTT ay >*†7 sa Therefore V2@?+z?)\(y+Z7)x+y+2) = xy(x+y)y/2(x? +?) +.2(x +2) (vy +z)4/2(x? +z?) z?(x—y)Ÿ x —y)?xyz w(x+y)(/2@2+y2)+zœx+y)+z)0+z)+Š c ») aoe IV
It suffices to prove that
Trang 2318 Solutions
We have
xy | (x+y) 4/2(x?2 +32) — Aap] > xz | (xt y)4/2(x2 +32) —4xy] - òồ
Hence, it suffices to prove that x—y)? x? — 10x 2) 2 f(z) =x l2) 2(x+y”) Ay + xy ety Than | — (21 oe ) +Z7(x+y) >0 Also, we have ƒ(z) - fb) = § (y—z) (21x? + 13° — 18xy— 8xz — 8yz) > 0 Therefore y(10x + 13y)(x—y)? > =x |(x+y)1/2(x? +7) —4x | — fle) > f0) =| (by) 262-49) Ay a 2x(x—y)P (+? +4xy) - r(0x+† 13x Sg (x+z)V 2ˆ +2) + tự 8(x+y) _ since
2x(x? +y? +4xy) y(10x + 13y) ` 2x(x? +7 + 4xy) _ #(10x+ l3y)
(x+z)v2@?+y?)+4o 8x1y) 2 13)+42w 8x+y)
_— 8xÌ+22xy-—I5xy?— I3yỶ >0
7 8(a+b)? —
Thus, our inequality is proved L]
Solution 5 (by Gabriel Dospinescu) We rewrite the inequality in the form
1 1 1\°
(a? +b”) (b* +07) (c? +a’) € +e +) >8(r +h +e)
a Cc
Setting a* + b* = 2x,b* +c” = 2y and c* +.a* = 2z, then x,y,z are the side lengths of a triangle and we may rewrite the inequality as 1 + 1 + 1 >—— x+y+z vy†z-xX vwZ†x—y vxdy—Z XZ By Holder Inequality, we have 2
xz(x+y+z) >3 xÌ(y+z—*), cyc
It suffices to show that
which is just Schur’s Inequality for fourth degree
Trang 2419 Problem 2.13 Let a,b,c be positive real numbers Prove the inequality (b+) (c+a) (a+b) a(b+c+2a) b(c+a+2b) c(a+b+2c) — Solution I After using AM-GM Inequality, it suffices to prove that (a+b)?(b+c)*(c+a)* > abc(a+b+2c)(b+e+2a)(e+a+2b)
Now, applying Cauchy Schwarz Inequality and AM-GM Inequality, we obtain
(b+e}(a+b)(a+e) > (b+6)? (a+ voc) — be ene vbe > be(Qa+b+c) 2 thtc Similarly, we have ab(a+b+2c)’, ca(c a 2ð)Ÿ (a+b)?(c+a)(c+b) > (c+a)?(b+c)(b+a) > Multiplying these inequalities and taking the square root, we get the result Equality holds if and only if a=b=c L]
Solution 2 We need to prove the inequality
(a+b)?(b+c)* (e+ a) > abe(a +b 4+ 20e)(b+e42a)(e+a+2b)
According to the AM-GM Inequality, we have
abc(a+b+2c)(b+¢+2a)(c+a+2b) IA ** abo(a+b+e)°
IA 64
a (ath +6) (ab + be + ca)"
And we deduce the problem to
64
(a+b)*(b+c)*(e+a)? > a (ath +6) (ab + be + ca)",
9(a+b)(b+c)(c+a) > 8(a+b+c)(ab+be+ca), ab(a+b) + bc(b + c) + ca(c+ a) > 6abc,
Trang 2520 Solutions Solution 3 (by Materazzi) We will try to write the inequality as the sum of squares, which shows that the original inequality is valid Indeed, we have
y (b+ce)? 3 y bre” a@at bre) paQatbre) ~~ Re aQat be) = (a+b+ 9 2BxrbrØ = (atb+ D> as 5 +c) To 5 °) | = (atb+ Ds lnon + = a) at 57 °) | = (atb+ Laat Frere 2)
which is obviously nonnegative
- Solufion 4 We have the following identity
4(b+e)? 2714 _ (7a+4b+4c)(2a—b-~ c) a(2a+b+c) a+b+c _—— ø(a+b+e)2a+b+c) — Ỷ
which yields that
(b+c)* „13 27 a
a(2a+b+c)— 4 4 atb+e It follows that
yet 2-2 a b ¢ )=3
xa2a+Ð+c) 4.4 a+b+e atb+e atb+o
Problem 2.14 Let a,b,c be positive real numbers Prove the inequality
(b+ce) (eta) (a+b)? >2 b+c c+a a+b
a(b+c+2a) Pb(c+a+2b) c(a+b+2c) — \b+c+2a ctat2b a+b+2c)°
Trang 2621
and
b+e cta a+b
a(2a+b+c) — b(2b+c+a) — c(2c+a+b)
Hence, we may apply the Chebyshev’s Inequality as follow b+e 1 b+e Flore 2a) aL : mm.) bi Cục a(2a +b+ c) 3 cye Cục = 0
This completes the proof Equality holds if and only ifa=b=c L] Solution 2 (by Honey_suck) We have a notice that
(b+ c)(b+c¢-—2a) (b+c)(c—a) (b+c)(a—b) >- a(2a+b+c) Ta To | _ (cta)(a—b) (b+c)(a—5) 7 |e Trai ¬ (a —b)? (2a? + 22 + c2 + 3ab + 3bc + 3ca) ore ab(2a+b+c)(2b+c+a) : cục cục
which is obviously nonnegative L]
Solution 3 Again, we notice that
(b+c)(b+c—2a) 3(b+c—2a) - (3a+2b+2c)(2a—b-~ c) — — >0 a(2a+b+c) 2(a+b+ec) 2a(a+b+c)(2a+b+c) — It follows that y Crib te~ 2a > By bten 2a _ 4 oe a2a+b+c) ~ 24 atbt+e 7
Problem 2.15 Leta,b,c be positive real numbers Prove that
(b+e}? (eta) (a+b)? s9( 4 4 b 4
Trang 2722 Solutions It suffices to show that a a 2) — +4) —_ > 6 wal È 2 p+e — 7 cye cye a 2a —— —— ——>
Lice t Liat bye —”
which 1s obviously true because a 2a 1 1 » oco te +, r2a+b+c —— Le ore b+c “Le + 2a+b | 4a > TT — nw c+ sul poe a — gy _* 2~22-1351.3c ` 8(a+b+e)? ~ Ya(2a+3b+ 3c) cye _ 4(a+b+e)2 —— #+b?+c2+3(ab-+ be+ ca) and
A(a+b+ce)? —3(a* +B? +7) — 9(ab+ be +a) = a +b? + —ab—be—ca> 0
Equality holds if and only ifa=b=c L] Problem 2.16 Leta,b,c be positive real numbers Prove that
ab+rPe+ed > (b+c—a)(e+a—b)\(at+b—-c)\(@+h+c)
Solution I By Schur’s Inequality for third degree, we have
Trang 2823 }›a(a—b)(a—c) Yala—b)(a—c) cyc cyc abe —_ œ@+j3+c3 (aÌ+bŸ+c`— abc))a(a—b)(a—c) > 0, cyc )
which 1s obviously true by AM-GM Inequality and Schur’s Inequality for third degree L] Solution 2 Without loss of generality, we may assume that a > b > c, then we have 2 cases
Case I If b+ c¢ <a, then the inequality is trivial since
(P+ec—a)(e+a—b)(a+b—c) <0
Case 2 If b+ c > a, then we have
(c+a—b)(a+b— c) a —(b—cy <a’,
bo—abc(b+c—a) = belbe+a(b+c—a)|(a—b)(a—c) > 0
It suffices to prove that
a’ (b> +c?) +a°be(b+c—a) >a’ (b+c—a)(a+h +c’), a(b’ +.3)+be(b+c-—a) > (b+e—a\(ae +b? +c?)
Now, since this inequality 1s homogeneous, we can assume that b+ c = | and put x = bc, then 1 > a> 5 and 5 >x > a(1-—a) The above inequality becomes
a(1— 3x) +x(1—a)* > (l—a)(a’ +1 —3x), f(x) = (4—8a+a’)x+at—a?+2a—1>0 We see that f(x) is a linear function of x, thus /6) > min} £5) flat—ay } But, we have a a— 2
/Ñ)- TT >0 and f(a(1—a)) =(2a—1)3 > 0
Thus, our proof is completed Equality holds if and only ifa=b=c L] Solution 3 (by nhocnhoc) By expanding, we see that the given inequality is equivalent to
Trang 2924 Solutions where Sa = bi+ch—3b?c*? +4+2abe(b+c), Sy = c?+a!—3c2a°+2abe(e+a), So = at+b*—3a?b? + 2abc(a+b) Without loss of generality, we may assume that a > b > c, then it is easy to check that S,,S, > 0 Moreover, we have SptS = 2a*+b4 —3e?b? +.c4 + 2abe(2a+b +e) — 3a? > 2a +b4-3e?b? +44 2ac?(2a+b+e)—3ac* = 2a1+b†—3a?b°+c°+ac(a+2b-+2ec > at +b —3e?b? = (a? — b*)(2a? — Bb”) > 0 It follows that
Sa(b—c)? +8)(e—aP +S8(a—by > S(e—a)?+S.(a—b)? > S,(e—a)? —Sp(a— by? =_ S;(b—c)(2a—b—c)>0
This completes the proof LÌ
Problem 2.17 7⁄a,b,c are positive real mumbers such that abc = Ì, show that we have the ƒollowing in- equality b c 3 +h +e >——+——+— b+c c+a a+b +2 2 Solution By GM-HM Inequality, we have that 4a 4b 4c 1 1 1 1 1 1 + + +6 < a(—+-)]+6(-+-]+c{[-+-]+6 b+e eta a+b b c ca a b
= a(b+c)+b?(c+a)+c? (a+b) +6abe 2a" (b+) +2b*(c+a) +2c? (a+b) 2ab(a+ b) + 2bc(b + c) + 2ca(e + a) 2(a +b?) +2(b +07) +2(c +.a°) 4(a +b? +03) IA IA Hence b 3 +R 48> b+c cta 44 atb 4 =, 2
Our proof is completed Equality holds if and only ifa@=b=c=1 Remark 1 We can prove that the stronger inequality holds
b 3
a+b +3—3> (6+4v2) c + + c _ =)
Trang 3025 Indeed, this inequality is equivalent to each of the following rt+hP+e b c 3 ore abe re’ 33> (644 ( + v2) = Tu*ya ca+b 3): _2 (a+b+e)(a—b)(a—©) cục (2a+b+c)(a—b)(a—c) abc > (3+2v2) (a+b)(b+c)(c+ta) ` X(a—b)(a—c)+Y(b—c)(b—a)+Z(c—a)(c—b) >0 where
XY = PP FING MOT NEFA _ (3.4 2) (2a+b+e)
eer rors) acres HỊ s (s+2v2) (2a+b-+e)
> eer rors) ee HH (s+2v3) (2a+b-+©)
= (2a+b+c) for eete -3-23| >0,
and Y,Z are similar
Now, assume that a > b > c, then we see that Z > Y > X, thus
X(a—b)(a—ec)+Y(b—c)(b— a)+Z(e— a)(c— b) Y(b—c)(b—a)+ Z(c— a)(e— b) Y(b—c)(b—a)+Y(c—a)(c— b) Y(b—c) >0
> >
Problem 2.18 Given nonnegative real numbers a,b,c such that ab + be + ca + abc = 4 Prove that
a +h? +c? 4+2(a+b+e)+3abe > 4(ab + be+ ca)
Trang 3126 Solutions which yields that p —4pq+9(4—q) > 0, or 3 ạ< p +36 4p+9 It follows that 34 36 pˆ+2p+12—-9g > p+2p4+12-9- 24 4p+9 _ Gptl8)4—p)(P—3) <9 4p+9 — This completes our proof Equality holds 1ƒ and only 1Ý ø = 5 =é= l ora=bồ=2,c= 0 and t1ts cyclic permutations LÌ Problem 2.19 Let a,b,c be real numbers with min {a, b,c} > 7 and ab + be + ca = 3 Prove that e+h+ce+49abe > 12 Solution Notice that from the given condition, we have 9 3 3 1
a> 167 7g (ab + be + ca) > eae te) > gio te)
It follows that 8a > b+ c Similarly, we have 8b > c+ aand 8c > a+b Now, using AM-GM Inequality, we obtain
Trang 3227 Hence, we may write the above inequality as X(a—b)(a—c) +Y(b—c)(b—a)+Z(c—a)(c— 5b) > 0, where - y= Mab + be+ea) | Con 64 (b—c) a+b+ec a+b+ec 2 _ 2ab+c)t(+2Ì và 2p 2c a+b+ec 6a? —b— _ a’ + (b+ c)(8a—b doo, a+b+ec and Y,Z are similar
We will now assume that a > 6 > c, then 4a(b b+c)* —4b — ? yoy — 443+©+(+©?=48(e+4)= (e2)? | gr py a+b+ec — 8(a—b)— (a—b)(a+b—2c) _ (a—b)(7a+7b + 10c) >0, a+b+ec a+b+ec It follows that X(a—b)(a—ec)+Y(b—c)(b—a)+Z(e—a)(c—b) > X(a—b)(a—c)+Y(b—c)(b— a) > Y(a—b)(a—c)+Y(b—c)(b- a) = Y(a-b)} >0
This completes our proof Equality holds if and only ifa=b=c=1 L] Problem 2.20 Let a,b,c be positive real numbers such that a*h? + b?c* +c?a* = 1 Prove that
(a? +b? +0*)? 4abey/ (a2 +b? +02)3 > 4
Solution According to AM-GM Inequality and Cauchy Schwarz Inequality, we have /(a? +b? +c?) > V3(2+ +c7)(a2?2 + b2c2 +c?a*)
= 4/3(a2+b?+c?)>atb+te
(a? +b? +c’)? +abe(at+b+c) >4,
It suffices to show that
a1+b†+c!+ abe(a+-b+e) > 2(aˆbŸ + bˆcˆ” +-c2aˆ), Yar(a —)(a—ec)+3 ab(a— b)”> 0,
cyc cyc
Trang 3328 Solutions
Problem 2.21 Show that ifa,b,c are positive real numbers, the following inequality holds
(a+b+c)*(ab+be+ca)* + (ab+be+ca)* > 4abe(a+b+e)>
Solution Setting x = +, y= tz = i, then we may write the above inequality in the form
(x-+y+z)? (xy+yztax)? +xy2(xty+z)? > 4(xyt+yzt+ax)
Using AM-GM Inequality, we have
xy2(x-++y+z)° > 3xyz(x+y+z)(xy+yz+z),
and we can deduce the inequality to
(x+y +2) (xy tyz+ 2x) + 3xy2(x+y+z) > 4(xy+yz+zv)”,
which can be easily simplied to
xy(x—y)”+yz—z)”-+z(z— x)” > 0,
which 1s obviously true and this completes our proof Equality holds if and only ifa =b=c L] Problem 2.22 Le a,b,c be real nưnbers from the imterval |3,4| Prove that
be ca
b
(a+b+e) ($: +) >3(a +h? +c’)
Solution I The given condition shows that a,b,c are the side lengths of a triangle, hence we may put a=y+z,b=z+x andc=x+y where x,y,z > 0 The above inequality becomes x+ x+d+z (x+y+z) y eters) > 30x? +y° +27 +xyt+yz+zr) cye y+Z We have (x-+y)(x +2) x(x+y)(x+z) x+y+z)|} —————| = } “— +} (w+y)(x+z (x+y de HE > M2 di y)(x+2) 2 X(x* + yz
cye VrzZ cye cye
From this, we may rewrite our inequality as 2 2 XX G432), y0 +2) ) Z0 T9) Z ZX AZ Xx ` 21212, y+z Z+x x+y x(x —y)(x —z) ,30—Z)W—*) „ZŒ—x)Œ—) >0, y+z z+x x+y
which 1s obviously true by Vornicu Schur Inequality
Trang 3429 Solution 2 (by nhocnhoc) We will rewite the inequality as
ah? +b? +a? —abe(a+b+c) ` 3(a° +b? +e”) —(at+b4+c)?
abe — a+b+c )
a 2 b 2 Cc 2
a ap) Oe) + ( — - = J (e-a)* + | —- = J (ab)? 20
ức ater) ‘ (3 71215) e2) HS aH) (aby 20,
Sa(b — ¢)? +Sp(c—a)? + S.(a—b)? > 0,
where 3 , › ›
a c
Sa= Fo Tho "Up apbpe’ > ob Gabpe
Without loss of generality, we may assume that a > 5 > c It is easy to see that Sz > S, > S;- Moreover, we have b 4 2 4 2(b — %+§e= =+-=—=———>“- ca ab atbt+ce7 a atbt+e _2U1c—4) vo a+b+c since b+c>4>a
It follows that S, > S, > 0 And we obtain
Sa(b—c)? +8)(e—aP +S8(a—by > S(e—a)?+S.(a—b)? > S,(e—a)? —Sp(a— by? =_ S;(b—c)(2a—b—c)>0
This completes the proof LÌ
Problem 2.23 Given ABC is a triangle Prove that
8 cos* A cos* Bcos” C +.cos24 cos2Bcos2C >0
Solution (by Honey_suck) Since cos? A+ cos* B+cos*C +2cosA cosBcosC = 1 andcosA cosBcosC < § , it follows that
3
cos’ A +cos”B+cos?C > T
and
(1 —cos”.A — cos” B— cos”)? = 4cos” A cos” Bcos”C
Trang 3530 Solutions It suffices to show that 4(a+b+e)3+9(a+b+e— 1)? 4(at+b+e—1)?+2(at+b+e)> (a+5+c~1J“+2(a+b+e) 3 1a+b+2 +1, 1 4p? +9(p—1)? 4p_-1)?+2p>-——— = ) +1, (p=a+b+c) _ _ 1)\2 34;~3)-~ ỦẺ cọ, 4p —
which is obviously true since p > ` L] Problem 2.24 Let a,b,c be positive real numbers such that a+b+c=3 and ab + bc + ca >
2 max {ab, bc, ca} Prove that
a?-+-bˆ+c > eh +h +a’
Solution The desired inequality is equivalent to each of the following inequalities
(a+b+e) (đˆ + lˆ +?) > 9(a2ˆ + bˆc? + cˆa`), 9(a*b? + b*c* +a") e+h+c _ 9(a*b? + bc? +-c7a’) a2 + b2 + c2 33(4ˆ°—ð2)(a”~— €) -Ð ————>2Y(a-5)(a—e) a2 -L b2 + c2 ox (at+b+c) > ) 3(a’ +b? +c’) > 3(a +b? +07) —(atb+c)’, X(a—b)(a—c) +Y(b—c)(b—a)+Z(c—a)(c— 5b) > 0, where be | 3(a+b)(a+c)—2(a@ +b? +7) 4+2(b-c)? = a@+43a(b+c)—be > ab+be+ca—2be ab + bc +ca—2max {ab, bc, ca} > 0, V
and Y,Z are similar