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

Old and New > toi a; Inequalities 111 and so nai TJ n—1 —_ mm a, Using this last inequality, it remains to | prove that +? Th a; +1 -S ny i=1 Tt đ | — @n41 nS i=1 Th =1 +đ„ +1 a, 71 ap —S a;+ (n— lay, + +1 —1 | - > Now, we will break this inequality into Gn+1 and (Tre +? +(n—-l)a)đxy 11 — đạiJ)>0 ?ì ben Tr — Sa) i=1 —An41 +1 ?ì s34 — Yer) i=1 > +1 Let us justify these two inequalities The first one is pretty obvious n n—L _ (n — 1)djp.1 — đu = II Io — „+1 + đn+1) + (nm — Yang, 21 +? —1 > Ong tangy: So (ai — Angi) +(n- n—-1 — any, = — Yan, - đa} =0 21 Now, let us prove the second inequality It can be written as nm Tr na; i=1 n Because n › 1=1 „+1 — nr À ad? n > dua nề ¿1 nr at n — À i=1 dị T,+—H i=1 n ae — › i=1 a~' > (using Chebyshev’s Inequality) and a,41 < —, it n is enough to prove that "nan + nr i=1 2+1 Tr + =1 yn but this one follows, because na??? + -a; n is proved 21 '> a; for all Thus, the inductive step 117 Prove that for any %1,%2, ,%n > with product 1, n SY) 1 2z [Som k=2 —Ín — va) n Clearly, we have x; < G and » x, > (n —1)G, so it suffices to prove that k=2 n n ono n (3) k=2 > 2G k=2 =.' k=2 We will prove that this inequality holds for all zs, ,„ > Because the inequality is homogeneous, it is enough to prove it when G = In this case, from the induction step we already have n n (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 k=2 k=2 - k=2 (z¿T— 1)” >0, clearly true Thus, we have proved that ƒ(#1,#as, ,#„) < ƒ(zi,G,G, ,G) plete the inductive step, we will prove that ƒ(zi,G,G, ,ŒG) r= Gr- Now, to com- < Because clearly ;> the last assertion reduces to proving that (n — 1) ((n-6? which comes down to + aos) n—2 Gonz +n 7" = > ((m-)6+ a) \* 2n — "Gn=3 and this one is an immediate consequence of the AM-GM Inequality Old and New Inequalities 113 118 | Gabriel Dospinescu | Find the minimum value of the expression » where @1,@2, ,@n, < 4142 ay m - Aan - add up to and n > is an integer Solution: We will prove that the minimal value is equality, we find that nm (m — 1)Soa? =: Indeed, using Suranyi’s nh— nm nm +NQ1a2 4n > Soap =1 In- => NA, 42 4n > S =1 a7 (1 — (n — 1)aj) =1 Now, let us observe that ¬ ¬ az (1— (n= 1)ax) =_ _ (i) = (aoa) and so, by an immediate application of Holder Inequality, we have » n n—-1 ; a — k=1 fol ; 1— (n— 1)ag But for n > 3, we can apply Jensen’s Inequality to deduce that nooo ¬ — k=1 n > mr | Dae — k=1 — n ¬ ns Thus, combining all these inequalities, we have proved the inequality for n > For n = 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@, < be nonnegative real numbers such that Qa ajtaz+ +a2 ——————ccc n > v3 — 114 Solutions Prove that ay ag l-da 4 an la > na IlTa2 — l1—a2 Solution: We proceed by induction Clearly, the inequality is trivial for n= Now we suppose that the inequality is valid for n = k —1, k > and will prove that for We Qk Lae > T14? aj+as+ +a; ka —_— “1-4? Is a2 Tá IV Qy lay k assume, without loss of generality, that aj > a2 > > ax, therefore a > az Using the notation w —= ap +a? + -+ap_, ———c k—1 from a > a, it follows that « > a> ay a2 I-aP 1d ¬ From „3 _ ga - #8 we obtain đpy —1 Tạp 1—z2 ———- k—I1 ` › mm k-1 (k — 1)(z — a) — (k — 1)z 1—z2 ` — 1—a2 › a7 k—I1` (a — ay)(a + ax) z+ and Qk 1—a? ka? — a3 — Thus, by the inductive hypothesis, we have TH Tay ca + (k —1)a l-r —(a—ag)(1+aaz) (1 — az)(1 - a?) ka 1l-@ — Qk (ra a — x (= = (k-1)(a —a)(1 + az) _ a= je (1 — x?)(1 - a?) 1— a2 1-a? a -5) eee) (1 — x?)(x +a) _ (a — ag)(œ — a¿)[—1 + #Ÿ + d; + ray + a(2 + ay) + 07 + ara,(x + ag) + a 2d] (1— a3)(1— a¿)(1— #?)(œ + a) Since 2? — az = ka” — d2 k(a — ag)(a + ag) — dị —= , it follows that k—1 k-1 (a — ag)(@ — ag) = k(a — ag)? (a + ag) (k-1)(a@+a,z) — _ ` Old and New Inequalities 115 and hence we have to show that x? + az + rap + a(# + ag) + aŸ + aag(œ + ag) + a zag > In order to show this inequality, we notice that ka? 2) ka? + (k — 2)a2k wet dy = k— “k—1 and z-+døyg >\x/z2+a})>a — Consequently, we have b _ VWk_1 z + q2 + zay + a(# + ag) + 8Ÿ + azag(% + ag) + a2zag > (z2 + a2) + a(# + ag) + a2 > k >>(sẫ: | k TT atte > 3a — 1, and the proof is complete Equality holds when a, = ag = = Gn Remarks From the final solution, we can easily see that the inequality is valid for the larger condition 14+, /2+4 m+—1 This is the largest range for a, because Íor đị = đa = - = đ„ạ + = # and ø„ = — Ì (therefore a = #\/ * n ), from the given inequality we get a > The special case n = 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 A0CLY < J 9abczz(a + b+ e)(œ + + 2), or (ab + bc + ca)(xy + yz 4+ 2x) > 36abcryz (2) From (1) and (2), we conclude that < (ab + be + ca)(xy + yz + 2x) > 36abcryz, therefore < 36abcryz To have = 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? + + z?) = Thus, it follows that > 36abcryz 121 | Gabriel Dospinescu | For a given n > 2, find the minimal value of the constant k,, such that if 71, %2, ,%, > have product VI+kzạzi + + VJV1tknze woe 1, then + — VI + kn#„ Thus, we can find a number M and some numbers a; > which add up to 1, such that 1+ 2n — | „ (ø—1)?" = Mag >n-1 Old Thus, az and New Inequalities 117 i and we have HSS r (areag~')] > (wea)