Tuyển tập đề thi vô địch bất đẳng thức thế giới P4

Old and New Inequalities 91 is trivial, because #¿_+ + 2(n — l)#¿ + 2441 < 2(n-1)- » x, for all i This shows k=1 that mp, > 2(n — 1) Taking x; = x’, the expression becomes + (n — 2)+ ert+ar-!4+2(n—-1) 142(n-l)et+a? and taking the limit when approaches + grt 14+2(n—-1)e™-!+xr-? 0, we find that m, ———~ < 2(n — 1) win and thus 2(n — 1) Now, we will prove that M, > ¬ Of course, it suffices to prove that for any L1,%2,- ,Ln > O we have n Lj 2—Z >" 99 Prove that if a,b,c are positive real numbers such that abc = 1, then 1 Ita+b Irb+e— l+c+a = Dba 1 245 2+¢ Bulgaria, 1997 Solution: Let x =a+6+c and y= ab+ bc+ ca Using brute-force, it is easy to see that 244 12+4 the left hand side is 4g ẻ, while the right hand side 1s aor tery ow, #2 + 2z + t + # + 4z + the inequality becomes wi t4e+y+3 1, z2+ 2z +-+zxụ 1214z+ _- — 9+ 4z~+2 2z + 3— zụ < 3-Yy z2 + 2z -+t+zụ — 9+4z~+2u' For the last inequality, we clear denominators hen using the Inequalitles # > 3,1 > 3,27 > 3y, we have s7 > 5z”, = > y?, vy? > 92, 52y > 15z,z > and z?u > 27 Summing up these inequalities, the desired inequality follows 100 [ Dung Tran Nam ] Find the minimum value of the expression — + atc a c where a,b,c are positive real numbers such that 21ab + 2be + 8ca < 12 Vietnam, First solution (by Dung Tran Nam): 2001 Let - = 2, y, — = z Then it is easy to check that the condition of the problem a, C becomes 2xyz > 2a” + + 7z And we need to minimize x + + z But z(2z — 7) > 2+ + 4ụ > 2z >7 x> 2x + 4y — 2z—7 94 Solutions Now, we transform the expression of the AM-GM 2z+4 : — = + — should vanish z +# -†+ z > #++ xy — 3z the numerator Inequality 14 so that after one application 11 „22+ — 11 et—+y-—+——£>274+—+42,/1+-— 2z 2x IS that2/1+ => x Second We 2œ — 7ï 2z But, it is immediate to prove x? and soz+y+z>5+"+— _> > 15 x We have equality for 15 Therefore, in the initial problem the answer is 21 achieved for solution: use the same substitution and reduce the problem to finding the minimum value of a+ y+ when 2xyz > 2a + 4y + 7z Applying the weighted the AM-GM Inequality we find that And also 2a + 4y + 7z > 105 - 125 515 -zð5 -y3 225 + + z)?2(2z + 4ụ + 7z) > > zi215 This means that (a+ ty: Because 2ayz > 2” + 4y + 7z, we will have (œ++z)> with equality for ¢ = 3,y = 2i2= 225 + =z++z2 15 > 101 [ Titu Andreescu, Gabriel Dospinescu ] Prove that for any x,y, z,a,b,c > such that xy + z + zz = 3, a b+e („+ ỹ z)+ b c+tu (z+z)+ +b +) >3 Solution: We will prove the inequality a b —(yt2)+——@t2)4 b+e C a+b (z +9) > V3(z + 9z+ z#) for any a,b,c, x,y,z Because the inequality is homogeneous in x,y, z we can assume that x+y+z= But then we can apply the Cauchy-Schwarz Inequality so that Old and New Inequalities 95 to obtain a b+e % cra! z + V3(z + z + z#) < Gab - (E5) v2:Eas Thus, we are left with the inequality ` ( ; ) +4+5< c re is equivalent to » craeznSä , which is trivial c+ø)(c Remark A stronger inequality is the following: b bac ——(U+z “+t c a) +o ;(#+) >3, V(+)( + 2) — (+ +2), which may be obtained by applying the Cauchy-Schwarz a +94 +2)4+ ytez 2+7 (a+b+2) ( beet +)= a+b ztrụ atb )~3z+u+z)> cha he b+e Inequality, as follows (Vytzt+Vvzt+a+Vr+y) 2(z++z)=ˆ V(œ +9)(œ + z) — (@++2) A good exercise for readers is to show that >+z++z+ y(at2z) So Vet V3(aw + 0z + z2) 102 Let a,b,c be positive real numbers Prove that (b+e— a)? (b+c)2+a? (cta-—b)? (c+a)?+2 | (a+b-c)? (a+b)+c? — 35° Japan, First solution: Let = + y= + cre ,2— b a= c gà Using the Cauchy-Schwarz b The inequality can be written — 1\2 UV Ù > z2+1 75 Inequality, we find that ye z2 +1 ` (z++z_—3) — z?+?+z?+3 1997 96 Solutions and so it is enough to prove that (c+yt+z2—-3) _ ?+uaz2ass But from Schur’s So ay > 2S a € Q/2”~—15),z+33 zy+18>0 Inequality, after (Sx) - et computations, we deduce that ay + 18> (Siz) -9Éz+ls>0, the last one being clearly true since Second some Thus, we have ` z > solution: Of course, we may take a+ 6+ c= The inequality becomes 4(1 — ø)? 27 awa TS ————a1+-aŠ10 But with the substitution 1-a = 2,1-—6=y,1—c = z, the inequality reduces to that from problem 47 103 [ Vasile Cirtoaje, Gabriel Dospinescu |] Prove that if a1, @2, ,@, > then a + đỂ + + + độ — > a1 +@2 +++ +an-1 Ín = 1) [ —_-— đụ, } where ứ„ 1s the least among the numDerS đ1, đạ, ; đạ, Solution: Let a; — ap, = 2; > for7€ {1,2, ,n —1} Now, let us look at n ) n a? —n a1 n |[& nr n—-l1 >a i=1 — (n— —— — Gy t=1 as a polynomial in a = ay It is in fact a+ (at a)" na ][(6 +) = (x= Ð ( ay tag t+++4+2n-1\" „— We will prove that the coefficient of a* is nonnegative for all k € {0,1, ,n—1}, because clearly the degree of this polynomial is at most n — For k = 0, this follows from the convexity of the function f(x) = +” n—-1 n—1 3z >(n—1) i=1 3i i=1 n—1 n Old and New Inequalities 97 For k > 0, the coefficient of a* is n—-1 n (2) › x; n—k — Tì ¿=1 › i14: - (212 _—p¬ 1