Tuyển tập đề thi vô địch bất đẳng thức thế giới P5

Trang 1

and so n—1 —_ nai TJ > toi 3 a; mm a, Using this last inequality, it remains to | prove that +? Th Tt Th +1 71 ny a; -S đ | — @n41 nS ap —S a, + i=1 i=1 =1 +1 —1 +đ„ +1 a; + (n— lay, - | > 0 Now, we will break this inequality into

Gn+1 (Tre +(n—-l)a )đxy 11 — đại J)>0 and +? ?ì Tr ?ì ben — Sa) —An41 s34 — Yer) > 0 i=1 +1 i=1 +1 Let us justify these two inequalities The first one is pretty obvious n n—L _ n—-1 Io (n — 1)djp.1 — đu = II — „+1 + đn+1) + (nm — Yang, — any, = 21 +? —1 — > Ong tangy: So (ai — Angi) +(n- Yan, - đa} 1 =0 21 Now, let us prove the second inequality It can be written as nm Tr nr nr „+1 n n T,+—H

na; — À ad? > dua nề at — À dị

Trang 2

Solution: Of course, the inequality can be written in the following form n n 2 n>-—(n-1) S2) + (so) k=1 k=1

We will prove this inequality by induction The case n = 2 is trivial Suppose the inequality is true for n — 1 and let us prove it for n Let

ƒ(#1,#a2, ,#p) = —(n — l) (2) + (so)

k=1 k=1

and let G = ”-⁄2a#3 #ạ, where we have already chosen x, = min{x,22, , 2p} It is easy to see that the inequality f(r1,%2, ,%n) < f(r1,G,G, ,G) is equivalent to nr n 2 +? (n — LS ax; — S2) > 2z [Som —Ín — va) k=2 k=2 k=2 n Clearly, we have x; < G and » x, > (n —1)G, so it suffices to prove that k=2 n n 2 n ono (3) > 2G =.' k=2 k=2 k=2

We will prove that this inequality holds for all zs, ,„ > 0 Because the inequality is homogeneous, it is enough to prove it when G = 1 In this case, from the induction step we already have n n 2 (ø—=2)3_z? +m—1> (yo) k=2 k=2 and so it suffices to prove that nm nm nm Soap tn-1>S 2x - (z¿T— 1)” >0, k=2 k=2 k=2 clearly true

Trang 3

118 | Gabriel Dospinescu | Find the minimum value of the expression » 4142 - Aan - ay m 1 1 add up to 1 and n > 2 is an integer where @1,@2, ,@n, < Solution: We will prove that the minimal value is =: Indeed, using Suranyi’s In- nh— equality, we find that nm nm nm (m — 1) Soa? +NQ1a2 4n > Soap => NA, 42 4n > S a7 (1 — (n — 1)aj) =1 =1 =1 Now, let us observe that ¬ 3 az (1— (n= 1)ax) = _ _ (i) ¬ = (aoa) and so, by an immediate application of Holder Inequality, we have n 4 3 » n—-1 ; k=1 ; 5 a — 3 1 2 fol 1— (n— 1)ag But for n > 3, we can apply Jensen’s Inequality to deduce that nooo n 3 ¬ | Dae — — 1 k=1 > k=1 — ¬ mr n ns

Thus, combining all these inequalities, we have proved the inequality for n > 3 For n = 3 it reduces to proving that

b

Viner 2"

which was proved in the solution of the problem 107

119 | Vasile Cirtoaje | Let a1,a2, ,a@, < 1 be nonnegative real numbers such that

ajtaz+ +a2 v3

Qa ——————ccc > —

Trang 4

Prove that ay 4 ag 4 4 an > na l-da la IlTa2 — l1—a2 Solution:

We proceed by induction Clearly, the inequality is trivial for n= 1 Now we suppose that the inequality is valid for n = k —1, k > 2 and will prove that Qy a2 Qk ka Lae > —_— lay Tá T14? “1-4? for aj+as+ +a; IV Is k 3

We assume, without loss of generality, that aj > a2 > > ax, therefore a > az Using the notation

ap +a? + -+ap_, ka? — a3 w —= ———c ca — ———- k—1 k—I1 ` 3 from a > a, it follows that ô > a> ơ Thus, by the inductive hypothesis, we have ay a2 đpy —1 (k — 1)z I-aP 1d TH Tạp — 1—z2 ` Tay 1—z2 — 1—a2 From › 5 5 › „3 _ ga - #8 mm a7 k-1 k—I1 ` we obtain (k — 1)(z — a) (a — ay)(a + ax) z+ and Qk (k —1)a ka — Qk a x a _ 1—a? + l-r 1l-@ (ra — = (= -5) 7

Trang 5

and hence we have to show that

x? + az + rap + a(# + ag) + aŸ + aag(œ + ag) + a zag > 1

In order to show this inequality, we notice that

ka? + (k — 2)a2 ka? 2) 2 k wet dy = k— 1 “k—1 z-+døyg >\x/z2+a})>a _ — b VWk_1 z + q2 + zay + a(# + ag) + 8Ÿ + azag(% + ag) + a2zag > (z2 + a2) + a(# + ag) + a2 > k | k > 1 TT 2 2 — >(sẫ: atte > 3a 1, and the proof is complete Equality holds when a, = ag = = Gn and Consequently, we have Remarks 1 From the final solution, we can easily see that the inequality is valid for the larger condition 1 14+, /2+4 m+—1 This is the largest range for a, because Íor đị = đa = - = đ„ạ + = # and ø„ = 0 — Ì (therefore a = #\/ * n ), from the given inequality we get a > 3

2 The special case n = 3 and a = v is a problem from Crux 2003

120 | Vasile Cirtoaje, Mircea Lascu | Let a,b,c, x,y,z be positive real numbers such that (atb+e\(e@tyt+z=(@P@4P4 ee)? +y42)=4 Prove that beryz < 3 A0CLY Z — J<ng Solution:

Using the AM-GM Inequality, we have

4(ab+be+ca)(xyt+yz+zx) = [(a+bt+c)? - (a +0? +c’)] [(ctyt+2) -(2? +y?+2)] =

Trang 6

therefore

(ab + bc + ca)(xy + yz + za) <1 (1)

By multiplying the well-known inequalities

(ab + be + ca)? > 3abc(at+b+c), (xu +ụz+ zz)) > 3z0z(w +ụ +2),

it follows that

(ab + be + ca)? (xy + yz + zx)* > 9abczz(a + b+ e)(œ + + 2), or

(ab + bc + ca)(xy + yz 4+ 2x) > 36abcryz (2) From (1) and (2), we conclude that

1 < (ab + be + ca)(xy + yz + 2x) > 36abcryz,

therefore 1 < 36abcryz

To have 1 = 36abcryz, the equalities (ab + bc + ca)? = 3abe(a + b + c) and (cytyz+zx)* = 3xyz(x+y+z) are necessary But these conditions imply a = b = ¢

and x = y = 2, which contradict the relations (a +6+c¢)(x2x+y+2) = (a@+6%4

c2)(z? + 2 + z?) = 4 Thus, it follows that 1 > 36abcryz

121 | Gabriel Dospinescu | For a given n > 2, find the minimal value of the

constant k,, such that if 71, %2, ,%, > 0 have product 1, then 1 + 1 + woe + — 1 VI+kzạzi VJV1tknze VI + kn#„ <mn — ] Mathlinks Contest Solution: 2 1 We will prove that k, = ao Taking #1 = #a = - = #„ = 1, we find that TL — 2n — 1 mà an So, it remains to show that (n — 1)? n 1 So cnt ra | te 2n — Ì -= ai (m—1)2 `

Trang 7

Thus, az <> HSS r (areag~')] > (wea) <1 (w=arg ~')- i and we have Now, denote 1 — (n — l)ag = by > O and observe that Shy = 1 Also, the above k=1 inequality becomes ¬ TIe-%> > (2n — 1)” (ITs -»] k=1 k=1 Because from the AM-GM Inequality we have II < (TH) k=1 our assumption leads to IIc — bự)” < PO be bp k=1 So, it is enough to prove that for any positive numbers a1, @2, ,@,, the inequality ˆ (n — 1)?" [[ (ai tae t+ +++ aK 1 tony t+ tan)” > Ty 2 «+ An (Qt Fda +++ +n)" k=1 holds This strong inequality will be proved by induction For n = 3, it follows from the fact that

He)’ E99) oss

Suppose the inequality is true for all systems of n numbers Let @1,a@2, ,@n41

abe

Trang 8

Now, using Huygens Inequality and the AM-GM Inequality, we find that 2n 2 2n 1+ An+1 > fat Gn+1 > (14 NAn+1 1 — 4a; n—-l1 Ila — aj) Tì wl and so we are left with the inequality an 3n+2 pq NOt > n” an+i(L+an+i) 1 n—1 ~ (n—1)??-(n+1)rt! n( + @n+41) = l+z, where # 1S n+1 1 if Gn4, > max{aj,a2, ,an} > — So, we can put n nonnegative So, the inequality becomes 2n 1 1 14—2_) ylttrt de n(x + 1) (x + 1)r7 Using Bernoulli Inequality, we find immediately that 2n Le x >3 n(œ + 1) e+1 Also, (1+ 2)"~! > 1+(n-—1)z and so it is enough to prove that 32 +1 ` l1+(n+ 1)z z-+l — I+(n-— l)z which 1s trivial

So, we have reached a contradiction assuming the inequality doesn’t hold for a certain system of n numbers with product 1, which shows that in fact the inequality

2n — 1

is true for k, = — and that this is the value asked by the problem

(n — 1)

Remark

For n = 3, we find an inequality stronger than a problem given in China Math- ematical Olympiad in 2003 Also, the case n = 3 represents a problem proposed by Vasile Cartoaje in Gazeta Matematica, Seria A

122 [ Vasile Cirtoaje, Gabriel Dospinescu | For a given n > 2, find the maximal value of the constant k„ such that for any %1,22, ,2%n > 0 for which 27 +23 + -+

x? = 1 we have the inequality

Trang 9

Solution:

We will prove that this constant is (,/n — 1)” Indeed, let a; = 2? We must find the minimal value of the expression nm [[a- va i=l nm [va i=l

when a; +a@2+ :+a@,, = 1 Let us observe that proving that this minimum is (.,/n—1)”

Trang 11

Glossary

(1) Abel’s Summation Formula

If a1, @9, ,@n, 61, b9, ,b, are real or complex numbers, and S;=a,+aot+ +a;,i = 1,2, ,n, then n n—1 ` a;Ö¡ = » Si(bi — bi41) + Sdn ¿=1 ¿=1

AM-GM (Arithmetic Mean-Geometric Mean) Inequality

If a 1, @2, ,@, are nonnegative real numbers, then

le 1

— So ai > (41 02 dn)”, nial

with equality if and only if a4, = ag = = ay This inequality is a special case of the Power Mean Inequality

Arithmetic Mean-Harmonic Mean (AM-HM) Inequality

If a1, @, ,@n, are positive real numbers, then

— ai ———_—

hi " L - _ 1

n

i=1 ay

with equality if and only if a, = a2 = = ay This inequality is a special case of the Power Mean Inequality

(4) Bernoulli’s Inequality

For any real numbers « > —1 and a > 1 we have (1+ 2)* > 1+ az

Trang 12

(5) Cauchy-Schwarz’s Inequality

For any real numbers a1, @2, ,@p, and bj, be, , bn

(a‡ + øã + + ø2)(bị + bộ + + b2) >

> (a,b; + asÖa + + a„,b„)Ÿ,

with equality if and only if a; and 6; are proportional, 7 = 1, 2, ,n

(6) Cauchy-Schwarz’s Inequality for integrals

If a,b are real numbers, a < b, and f,g : [a,b] > R are integrable functions, then ( / “Heltei) < ( / Pee) ( / rey) , (7) Chebyshev’s Inequality If ay < ag < < ay and 01, bo, ,b, are real numbers, then ?ì 1 Tt mr I)If b: < by < < by then Said; > — (Soa; } bị]: “1 „=1 „=1

2)If bị > bạ > > bạ then Ð ` a¡b¡ <

(3) Chebyshev?s Inequality for integrals

pr

SI

If a,b are real numbers, a < 6, and f,g: [a,b] > R are integrable functions and having the same monotonicity, then b (b—a) [ Ƒ(z)g(z)dz > [ sou | g(z)dz and if one is increasing, while the other is decreasing the reversed inequality is true (9) Convex function

A real-valued function f defined on an interval J of real numbers is convex if, for any x,y in J and any nonnegative numbers a, with sum 1,

flax + By) < af(x) + Bf(y)

(10) Convexity

A function f(x) is concave up (down) on [a,b] C R if f(x) lies under (over) the line connecting (a1, f(a1)) and (6), f(61)) for all

ax<a<2<b, <b

A function g(x) is concave up (down) on the Euclidean plane if it is concave

Trang 13

(14)

with R

Concave up and down functions are also called convex and concave, re- spectively

If f is concave up on an interval [a,b] and Ai, A2, ,An are nonnegative numbers wit sum equal to 1, then

Ai f (#1) + Aof (42) tot Ànƒ(#n) > ƒOlzi + Às#¿ + + Àn#n)

for any %1,2%2, ,%p in the interval [a, b| TẾ the function is concave down, the inequality is reversed This is Jensen’s Inequality

Cyclic Sum

Let n be a positive integer Given a function f of n variables, define the cyclic sum of variables (21, #a, , #„) aS Sf (1,3, «+5 2n) = f(x1,%2, ,n) + (Z3, 3, ,®n; #1) cực + + ƒ(@„, đ1, đa, , Gp—1)- Holder’s Inequality 1 If r,s are positive real numbers such that — + — = 1, then for any positive rs real numbers a1, G2, -,@n, b,,bo, ,bn, 1 Ss 3 3 4 n n 7=1 < 7=1 = - i=l nr a Tì Tì Huygens Inequality

lÍ p1,7Ø», ,Ðn,01,05, ,0n,Ðt,ba, ,bạ are positive real numbers with

pit pot + pn = 1, then

n n n

Tú: +ủ¿)?: > II + II:

21 21

=1

Mac Laurin’s Inequality

For any positive real numbers 21, %2, -.,%n,

Trang 14

(17) (19) where › địa Vig (+ - * Li, kị 1<ii<i2< <iy<n n k For any real number r > 1 and any positive real numbers Sp = Minkowski’s Inequality Q1,@2, -,@n,61,69, ,bn, pa + nr) < (>: a) + (>: n) |

Power Mean Inequality

For any positive real numbers a 1,@2, ,@, with sum equal to 1, and

any positive real numbers 21,%2, ,%, we define M, = (ay;a{ + agr5 + £ ant")? if r is a non-zero real and My = 22%? 0%, Moo =

maxr{x1,%2, ,2n}, M_o = min{x1, £2, ,£n} Then for any reals s < t

we have M_,, < M, < Mi < My

Root Mean Square Inequality

If a 1, @2, ,@, are nonnegative real numbers, then

la? + gộ + + q2 s01 + đa + + đa

n — n ,

with equality if and only if a; = ag = = Gn

Schur’s Inequality

For any positive real numbers x,y, z and any r? > 0,#”(œ—)(—z)+7(u— z)\(y—2x2)+2"(z-—2)(z—y) > 0 The most common case is r = 1, which has

the following equivalent forms:

Trang 15

(20) Turkevici’s Inequality

For any positive real numbers #, 1, 2, Ý,

aityt tts tt 2xuzt > 2u? + g2z2 + z2? 4 Pe ee? ty

(21) Weighted AM-GM Inequality

For any nonnegative real numbers @1, G9, ., dn, if w1, we, ., Wy, are nonneg-

ative real numbers (weights) with sum 1, then

101 _ Ww ne

Trang 17

Định dạng
Số trang	17
Dung lượng	531,97 KB