Old and New Inequalities 61 First solution: Using the identity (a + 6)(6+ c)(c +a) = (a+6+4+c)(ab+ be + ca) — we reduce the problem to the following one ab + be + ca + — > — atb+e Now, we can apply the AM-GM b+6 > Inequality in the following form ,/ (ab + be + ca)? ————— an ——— 9(a+b+) And so it is enough to prove that (ab + be+ ca)? > 9(at+b+c) But this is easy, because we clearly have ab + be + ca > and (ab+ be + ca)? > 3abc(a+ b + e) = 3(a+b+ ©) Second solution: We will use the fact that (a + b)(b+ c)(e +a) > gia + b+e)(ab + be + ca) So, it is enough to prove that (0b + be + ca) + ; Inequality, we can write a+b+c > Using the AM-GM „| (œb + be + ca)? >1 8l(a+b+ec) — g (ab + be+ ca) + ————— a+b+c7— ? because (ab + be + ca)2 > 3abe(a + b + e) = 3(a+ b+ e) 57 Prove that for any a,b,c > 0, (a7 + b` + c?)(a+b— e)(b+e— a)(e+a— b) < abe(ab + be + ca) Solution: Clearly, if one of the factors in the left-hand side is negative, we are done So, we may assume that a,b,c are the side lenghts of a triangle ABC With the usual notations in a triangle, the inequality becomes (a? +b? +c’) atb+e < abc(ab+be+ca) & (at+b+c)(ab+bc+ca)R? > abc(a* +b? +c’) But this follows from the fact that (a + 6+ c)(ab + bc + ca) > 9abc and 0< OH? =9R?-— a2 — b — cẺ 62 Solutions 58 | D.P.Mavlo | Let a,b,c > Prove that 161 Sal fy gle We 1)(b Mes+ Mer) 1)(c Dery)+ Btatbtesn41 rere b b + abe Kvant, 1988 Solution: The inequality is equivalent to the following one La — eatery or abe Ya+ — — +) Dato _>939=————“^= a c+ abe + a > 2(Soa+ am) But this follows from the inequalities 2 + ọa > 2ca œ“bc + Pb > 2ab,“ca + —C > 2bc,c“ab a and a-c+ — > 2a,b°a+ — > 2b,c°b+ = > 2c e a b 59 [| Gabriel Dospinescu | Prove that for any positive real numbers 21, 2%2, ,2n with product we have the inequality n n”-][(@P+l> al i n n i=l i=1 2) 3] n ? Solution: Using the AM-GM xt1 1+z7 and l+z‡ Thus, we have Inequality, we deduce that 12 l+zỹ + —— l+zỹ fon Len—I1 1+z„¡ l+z? > n Old and New Inequalities 63 Of course, this is true for any other variable, so we can add all these inequalities to obtain that which is the desired inequality 60 Let a,b,c,d > such that ø + b+ e= Prove that a® +6° + + abed > 1 { 5, stat: Kvant, 1993 Solution: Suppose the inequality is false Then we have, taking into account that abc < 2m 1 the inequality d (5 — abe) >@+P43 - We may assume that abe < Now, we will reach a contradiction proving that a® + 6? + c? + abed > T It is sufficient to prove that a+h+e—— ———T—————— abc But this inequality is equivalent to » a3 = 3abe+1—3 aửe +a3+b”+c7> + l5abc > + We use now the identity 1+ 9ab ` ab and reduce the problem to proving that » ab< _ which is Schur’s Inequality 61 Prove that for any real numbers a, b,c we have the inequality +22)? +b2)?(ø— e)?(b— ø)” > (1+a”)(1+ð2)(1+c”)(a— b)”(b— e)”(e— a)° AMM Solution: The inequality can be also written as » 52 (1 ren? > (of course, we may assume that a,b,c are distinct) Now, adding the inequalities (1+ a?)(1 + 07) (+6)d+e) J, (1+c?)(a—b)2 — (1+a@?)(6-c)? ~ (which can be found using the AM-GM Inequality) we deduce that (1+ a7)(1+ 0?) 2>axzza- 14+ 06? |a— b|le— Ù| >> 14+ 6? I6-s0-a 64 Solutions and so it is enough to prove that the last quantity is at least But it follows from » 1+? jy |l(b— a)(b—e)|— SS + - = —c) and the problem is solved 62 | Titu Andreescu, Mircea Lascu | Let a, 2, y,z be positive real numbers such that xyz = and a > Prove that re ụ2 Uu+zZ z+z + z2 ety First solution: We may of course assume that x > y > z Then we have x y Z +2Z_— and 2%! z1+#z E+Y > y?-! > 2-1, Using Chobystiew's E23 (Le) Inequality we infer that ee ì (LS)% Now, all we have to is to observe that this follows from the inequalities » re1 >3 (from the AM-GM Second Inequality) „3 and s~— „-+z—2 solution: According to the Cauchy-Schwarz (lu+2)+u(e+2)+2(e+y)] ( Inequality, we have: a + +5) > (ý 2+2 +z®) Thus it remains to show that (0 +y +2)" > 3(ay + yz + 20) Since (x + y + z)* > 3(zy + yz + zz), it is enough to prove that lta ite ita +? +27 +22 >r+ytz From Bernoulli’s Inequality, we get of =[L+(z—-U|9° >1++#Z¿-n= Ita 1— c4, 14 and, similarly, 1+a 1+a La 3(1-a) lta (c+y+z)-(@+y4+z)= ` Old = — Equality holds for # = and New Inequalities 65 — (e+w+z~3) > “—(3ÿwz— 3) =0 = z = Remark: Using the substitution = a+1(6> 2) andz= -, y= Đ z= - (abc = 1) the inequality becomes as follows a? (b 1 +c) b2(e + a) đ*{(a+ b) 2 For = 3, we obtain one of the problems of IMO 1995 (proposed by Russia) 63 Prove that for any real numbers 21, ,2%n,Y1, -,Yn Such that 2?+ -+22 = yee dyad, ?ì (Z1a — #s1)” < ( — > een k=1 Korea, 2001 Solution: We clearly have the inequality (amram YS wan? = (Set) (Sout)- (Sam) = - (1- Soa) mm) 1—a 7=1 Also, from the above relations we can see imme- we have equality if and only if a,,a2, ,a@, is a permutation of the 1,2, ,n 65 [ Calin Popa ] Let a,b,c be positive real numbers such that a+b+c¢= Prove that b/c a(V3c+vab) + ca + b(V3a+vbe) avb ` 3v3 c(V3b+ ca) — - Solution: Rewrite the inequality in the form bc > 3v3 ca + ——+a b With the substitution ô = be ca ab 4/—,y = „—'®=Ứ= the conditlon a + b-+e= a c becomes xy + yz + zx = and the inequality turns into # V3y But, by applying the Cauchy-Schwarz » a V3ay + ayz > > 3v “4 + yz Inequality we obtain » 7) V34 3xryz > ty V3 + i v3 where we used the inequalities (Siz) >3 (So zy) 3v3 =—,4 and xyz < wa Old and New Inequalities 67 66 | Titu Andreescu, Gabriel Dospinescu | Let a, 6, c,d be real numbers such that (1 + a?)(1 + 67)(1 +c?)(1 + d?) = 16 Prove that —3 < ab+bce+cd+da+ac+ bd — abcd < Solution: Let us write the condition in the form 16 = H¿ +a)- Ta —i) Using symmetric sums, we can write this as follows 16= (1-10 a- Sab +i > abe + abcd) (1+% a- Sabi D abe + abcd) So, we have the identity 16 = (1 — Ð` ab + abcd)? + (SS a— >> abc)” This means that |1 — S> ab+ abcd| < and from here the conclusion follows 67 Prove that (a? + 2)(b + 2)(c? + 2) > 9(ab + be + ca) for any positive real numbers a, 6, c APMO, 2004 First solution: We will prove even more: (a? + 2)(Ù2 + 2)(c? +2) > 3(œ + b + c)? Because (x+b+e) < (la| + |b| + |c|)”, we may assume that a,b,c are nonnegative We will use the fact that if x and y have the same sign then (1+ 2)(1+y) >1+2+y So, we write the inequality in the form (+2) > othe and we have three cases i) If a,b,c are at least 1, then TT (“ 2_ +1) >14505 2_ ( a) » 2) ii) If two of the three numbers are at least 1, let them be a and 6, then we have a-1., peal IV Il Pal +2 (a? +b? +1) (1?+1% +c’) ` (a+b+c) by the Cauchy-Schwarz iii) If all three numbers are at most 1, then by Bernoulli H(G and the proof 1s complete — Inequality +) c2 Inequality we have 68 Solutions Second solution: Expanding everything, we reduce the problem to proving that (abc)? + 23 `a?0? +45 0a? +8> Because inequality 3300? > (abc)? + » `ab and a?+2> 23 `a?0? +6> 2À ` ab 9S — ab 4À ` ab, we are left with the Of course, we can assume that a,b,c are non-negative and we can write a = 27,b = y’,c = z” In this case 2S ab-S oa? =(z++z)(z+— z)(u+z— z)(z+z— 9) It is clear that if x,y,z is trivial Otherwise, are not side lengths of a triangle, then the inequality we can take x =u+v,y =v+w,z = w-+u and reduce the inequality to ((u+v)(v+w)(w+u))*+2> 16( + + u)uU We have ((u+v)(v + w)(w +u))* +141>3%/(ut vs (vtw)t(utw)! it remains to prove that the last quantity is at least 16(u+v+w)uvw down to 16 (utv)i(v+w)*(w+u)* and This comes > sp (wap) (u ++u) But this follows from the known inequalities (utv)(vt+w)(w+u) > gute +w)(uv + vw + wu), (uv + vw + wu)* > 34(uow)3, Third wu + +~+ > 3u) solution: In the same manner as in the Second solution, we reduce the problem to proving that (abe)? +2> 23 `ab— » Now, using Schur’s Inequality, we infer that 9ab 23 2ab—À `a? 3/(abe), the problem is solved But the last assertion follows from the AM-GM Inequality Old and New Inequalities 68 | Vasile Cirtoaje | Prove that IÍ < z < a) (1— zø)(1— z)(1— zz) 0; 69 < z and z+ + z = zz +2, then b) Jays ay < 1a’y? lay" < = 55 Solution: a) We have (l—ay)(1—yz) =1l-—ay—yz+ay?2z =1—a2y—yz+y(@t+y+2—-2) =(y-1* >0 and similarly (1—øz)—zz) =(1—z)“>0, So the expressions — xy, — z (1—zz)(1— zu) = (L— ø) >0 and — zz have the same sign b) We rewrite the relation x+y+z = xyz+2 as (1—x)(1-y)+(1-z)(1—-zy) = Ifz > then z > # > z > and so (I — z)(1 — ø) + (1— z)(1— zg) > 0, impossible So we have x < Next we distinguish two cases 1) xy < 1; 2) zy > I1 32 1) ty (z0 — l)U, (@T— 1)(2—#— z) > so 2> z(1+ 0) Using the AM-GM Inequality, we have 1+ > 2Vÿ and 1+ 2>3z = = 1+ 548 5301-32 Thus we have > 2e,/y and 32 which means that xy < and xy? < 32 The equality x?y = takes place when x = y = and the equality x?y”? = 2z takes place when x = 3° y= r=2 69 | Titu Andreescu | Let a, b,c be positive real numbers such that a+b+c > abc Prove that at least two of the inequalities 2 a Ob eg Cc 78 b ¢c¢ 4a 8g 28 ec a Log ob are true TST 2001, USA Solution: The most natural idea is to male the substitution -— = x, ; = Y, : = z Thus, we a have #,,z > Ö and z# + z + zz > and we have to prove that at least two of the inequalities 2z+3-+6z > 6, 2u+3z-+6z > 6, 2z+3z~+6y > are true Suppose this is not the case Then we may assume that 2z-+ 3„-+6z < and 2z+3z-+ổy < Adding, 5-5 we find that 5a+9y+8z < 12 But we have x > + ễ, Thus, 12 > +9 +8z Ù # + z 70 Solutions which is the same as 12(y+z) > 5+9y?+82?+ 1l2yz @ (2z—-1)?4+ (8y+2z—-2)? < 0, which is clearly impossible Thus, the conclusion follows 70 | Gabriel Dospinescu, Marian Tetiva ] Let x,y,z > such that z + Prove that +Z—+xUz (— 1)(w— 1)(z— 1) < 63 - 10 First solution: Because of x < xyz > yz > (and the similar relations xz > 1, zy > 1) at most one of the three numbers can be less than In any of these cases (2 < 1, y > 1, z > or the similar ones) the inequality to prove is clear The only case we still have to analyse is that when x > 1, y >1 and z> In this situation denote x-l=a,y-—1=6,z-l=c Then a,b,c are nonnegative real numbers and, because zœ=a+l,=b+l,z=c+], they satisfy a+1+b6+1+c+1=(a4+1)(b4+1) (c+), which means abe + ab+act+ be = Now let « = Wabc; we have ab + ac + be > 3V abache = 32’, that’s why we get a + 34? 10 — Mathematicorum Old and New Inequalities 77 with x,y, 2,t,u > It is clear that + atabe — y tt tsi 1+ ab + abcd ~ Ta T+ + Z U tạ 1 Writing the other relations as well, and denoting — = a,, — = aa, T— = đa, TT x y Z as, we have to prove that if a; > 0, then ` _ 0a 1d a, Using the Cauchy-Schwarz +agt+a5 a4,-= u „1.1 - Inequality, we minor the left-hand side with 4S? 2S% — (ag + a4)? — (a, + a4)? — (a3 + a5)? — (ag + a5)? — (a, + a3)?’ where S = So By applying the Cauchy-Schwarz Inequality again for the denominator Of the fraction, we obtain the conclusion 78 | Titu Andreescu | Prove that for any a,b,c, € (0, 3) the following inequality holds sỉn ø - sin(œ — Ö) - sin(œ —e) sin(b + c) sinb-sin(6 — e) - sin(b — ø)_ sin(c + a) sinc- sin(e — a) - sin(e — b) >0 sin(a + b) — TST 2003, USA Solution: Let x = sina, y = sinb, z = sinc Then we have x,y,z > It is easy to see that the following relations are true: sina - sin(a — b)- sin(a — c) - sin(a + 6) « sin(a +c) = x(a? — y?) (a? — z”) Using similar relations for the other terms, we have to prove that: > 2(2? — y°)(a? —y*) > With the substitution = /u, y = Vv, z = Vw the inequality becomes » Vu(u — v)(u — w) > But this follows from Schur’s Inequality 79 Prove that if a,b,c are positive real numbers then, Va4 + b1 + c+ + Veh? + Pet 2a? > Va3b+ Bet Gat KMO Summer Program Solution: It is clear that it suffices to prove the following inequalities So at +S oa? b? > S > ab + áp? Vab? + be? + ca Test, 2001 78 Solutions and (Soot) (Shas?) = (Sad) (2á) The first one follows from Schur’s Inequality So at+abeS a > S a?b+ áp and the fact that Sab? > abe S~ a The second one is a simple consequence of the Cauchy-Schwarz Inequality: (ab + Betcha) < (ab? + b?c? +? a7)(at + bt 4+ &*) (abŸ + beŠ + ca®)? < (a2bŠ + b°e? + e?a?)(a* + b + €) 80 | Gabriel Dospinescu, Mircea Lascu | Eor a given ø% > find the smallest constant k, with the property: if a1, ,@, > have product 1, then a1 Q2 4203 (a? +a2)(az +a) 4 Andy, (a3 +.a3)(a? + a2) Qe2r-1 "= amETTTj@TT+T) for all z > Clearly, this implies k, + m„ — ”(+z)” ` 7+ — ”1+ø > ?— Let us prove that ø — 1s a good constant and the problem will be solved First, we will prove that (2? + y)(y? + ©) > zw(1 + z)(1 + g) Indeed, this is the same as (x + y)(x — y)” > So, it suffices to prove that (1+a1)(L+az) + 1 (1+ a2)(1 +43) Now, we take a, = „HỘ c.«a đựy — “ #2 »“ k=1 2-1 R+1ER+2 (Z + #+1)(#k+1 + #&+2) ) >9, which can be also written in the following from n Lk » hol Lk + Le41 + #11 (tp + #g+1)(#e+1 + #g+3) > Clearly, nm nm xem < n-2 (1+ an)(1+ ai) ~ and the above inequality becomes #1 (1 _ fess + —— ———————_| nee "+ 2n Old and New Inequalities 79 So, we have to prove the inequality n S- Chee >1, Mek + €p41) (p41 Using the Cauchy-Schwarz + Le+2) Inequality, we infer that n n +1 2„ k=1 (tp + #g+1)(#g+q de k+1 > + #eg+2) — xẽ (tp + Le41)(Le+1 + Le42) k=1 and so it suffices to prove that n » Th «| > Sox; k=1 nr Tr +23 k=1 7i + 0a k=1 k=1 But this one is equivalent to n › Liv; > › n %g#k+1 + ) 3g'k 1 at+b+c)(p+q+r), Wl aptbgter+a° +b +e =(atb+e\(ptqtr) Since 4(a+b+e)(p+qg+r) 1989 =[(at+p) + (b+ q) + (ctr) S0 Solutions 82 [ Vasile Cirtoaje | Prove that the sides a, b, c of a triangle satisfy the inequality 3(0 4242-1) b ¢ oa First 50(2 4622) a b oe solution: We may assume that c is the smallest among a, b,c Then let x = b— at c After some computations, the inequality becomes (3a—2e)z°+ (ø +e- 1) (a—c)? > © (3ø—2e)(2b—a—e)?+(4b+2e—3a)(a—e)Ÿ > which follows immediately from 3ø > 2c, 4b + 2e — 3ø = 3(b+e— ø)+b—ec>0 Second solution: Make the classical substitution a =y+2,b=z++2,c=2x+y and clear denomi- nators The problem reduces to proving that #2 + + z3 + 2(z?u + )z + z +) > 3(z2 + uz + zz”) We can of course assume y=xt+m,z=2+Nn that x is the smallest among x,y,z Then we can write with nonnegatives m and n A short computation shows that the inequality reduces to 2x(m? — mn +n?) + m3 +n? + 2m?n — 3n?m > All we need to prove is that m? +n? + 2m?n > 3n?m | (n—m)> —(n—m)m? +m? > and this follows immediately from the inequality t? + > 3, true for t > —1 83 | Walther Janous | Let n > and let 71, 22, ,2%, > add up to Prove that Crux First Mathematicorum solution: The most natural idea is to use the fact that n— x; 1-2; n—l Byte te + R-pt@U@yi t+ + fy Thus, we have (i=) i=l Ly =1 2E (: +— „ #122 s.-2—19¿L1 -Jn ) ... (utv)i(v+w)*(w+u)* and This comes > sp (wap) (u ++u) But this follows from the known inequalities (utv)(vt+w)(w+u) > gute +w)(uv + vw + wu), (uv + vw + wu)* > 34(uow)3, Third wu + +~+ > 3u) solution:... ca) and > [« -%| Inequality, we reduce this inequality to the following one (3) >|H«=9| This one is easy Just observe that we can assume that a > > ¢ and in this case it becomes (œ — b)(a — e)(b... 9, or, with this expression of z, day" — ay — (ex +y) xy—**—1