Old When and New Inequalities 31 we have equality? Gazeta Matematica Solution: First, we write the inequality in the following form (1+ 32) (+3) (+) (+Š) But this follows immediately from Huygens >7, Inequality We have equality for r=2y=s.2=1, 11 | Mihai Piticari, Dan Popescu | Prove that 5(a? +b? +0”) < 6(a? +b? +c?) +1, for all a,b,c> O witha+6+c=1 Solution: Because ø+ b+ e = 1, we have a + b3 + cŸ = 3abe + a2 + b + cŸ — ab — be — ca The inequality becomes ÿ 18abe+ 6(a2 + bŸ + c?) — 6(ab + be + ca) + © 18abe + — 2(ab + be + ca) + > 6(ab + bc + ca) ÿ and a> < 2 i€ {1,2, ,n} such that x, + T Prove that x; € 0, | tr all n Solution: Using the Cauchy-Schwarz Inequality, we get (ø—#i)” 2Vab and 6+ c > 2Vbc ca) > Writing the other two inequalities and adding them up give the desired result 14 For positive real numbers a,b,c such that abc < 1, prove that a b + + >a+b+e boc First oa solution: If ab+ be +ca < a+6+c, then the Cauchy-Schwarz Inequality solves the problem: (x+b+e€) a boc ¢_ oa @etPateb abc — 1 athe abc Satbete — 1S Otherwise, the same inequality gives oe Cc _ Werbared, (ab + bc + ca)? atbte abc abe OG >atb+e (here we have used the fact that abc < 1) Second solution: Replacing a, b,c by ta, tb, te with t = = preserves the value of the quantity in the left-hand side of the inequality and increases the value of the right-hand side and makes at - bt - ct = abct? = Hence we may assume without loss of generality that abc = Then there exist positive real numbers 2, y, z such that a = x b= ¬ c= `, The Rearrangement ' Inequality gives + ty? +23 > xy ° +y?zt+ 27a ° Old ‘Thus boc and New wv +y?423 ca Inequalities 33 ` z2 + 2z + z?œ 272 LYZ =øa+b+ec as desired Third solution: Using the AM-GM Inequality, we deduce that 2a b 3/ a" —T +~ >ở\| — > 3a Den be = 84 2b Similarly, —+ < > 3b and —+ c a a > 3c Adding these three inequalities, the conclusion is immediate Forth solution: Let # = fab* \ y= [cat ' w= | be* a Consequently, a = xy?,b = zx7,¢ = yz’, and also xyz < Thus, using the Rearrangement Inequality, we find that 3>” >”n` "nh a z? ` oe ay 15 [ Vasile Cirtoaje, Mircea Lascu | Let a,b,c,2,y,z be positive real numbers such thata+a>b+y>c+zanda+b+c=2+y+2z Prove that ay+bzr > ac+2z Solution: We have ayt+bx—ac—xz = a(y—c)+x(b—z) = a(a+b—x—-z)4+2(b-z) = a(a—x)+(a+2z)(b—z) = s 1 » = s(a—#)” + s (47 —#z”)+(œ+z)(b— z) = s(aT—#)” + s(a + #)(ø — # + 2b — 22) 1 The equality occurs when a= 2, b=2,c=yand 2x >y+42z 16 | Vasile Cirtoaje, Mircea Lascu | Let a,b,c be positive real numbers abc = Prove that 1+ Solution: We set © = equivalent to = a 1+ ` so that a+b+e™7 -,y a = abtac+be Junior TST 2003, Romania z = — and é observe that + + Z2 + z2 ` > xyz z++z = The inequality is 34 Solutions From (x + y +z)? > 3(ay + yz + zx) we get ————">~ + z + zz > + Ta" (cty+z) so it suffices to prove that 1+ (c+ytz)? > The last inequality is equivalent to ~a+yt+z (1 — ==) z++z > and this ends the proof 17 Let a,b,c be positive real numbers Prove that a3 bồ c aa b c2 JBMO First We 2002 Shortlist solution: have g3 ao > ta— beat +b > ablat) & (a—b)*(a +b) > 0, which is clearly true Writing the analogous inequalities and adding them up gives eo bồ @ b2 c2 2b? —=+=š=+—>>_— > —=—+—>+— B8 g2 —b+—+ủ-— TT cr 7— re-a be a Second solution: By the Cauchy-Schwarz Inequality we have a3 b3 C3 p2 a2 b2 c 2\ —4+—~+4+—}]>/[/—4+—4— (a+ b +o (B+5+S)2(F +—+ ) : so we only have to prove that —=+T+>a+b+e b Cc a But this follows immediately from the Cauchy-Schwarz Inequality 18 Prove that ifn > and %1,%2, ,%, > have product 1, then l+za + #12 + 1+ 222+ %2%3 fee + ——— _ 1+2%,+%n21 >] Russia, Solution: - su gs 2004 a2 a3 a1 Ớ1 a2 an We use a similar form of the classical substitution 7; = —,r%q = —, ,%, = — In this case the inequality becomes Ớ1 đ1 + đa + 0a + a2 đ2 + 0a + G4 an + -+—————>] Gyn + A, + Qe Old and New Inequalities 35 and it is clear, because n > and a; + aj41 + aj42 < a1 + ao +-+-+ a, for all 19 [| Marian Tetiva | Let 2, , z be positive real numbers satisfying the condition a? ty a) 8Ụ2 € , b)z++z< | Co Prove that +27 4+ 2Qaeyz = ay?2 ? < se —8 b) Clearly, we must have x,y,z € (0,1) If we put s = z+-+z, we get Immediately from the given relation s?—284+1=2(1-2)(1-y)(1-2z) Then, again by the AM-GM obtain fa asti0‹€©z+— 2z < | NO that x,y< 5" Because of symmetry, we may assume 36 Solutions On the other hand, = a 4y? +274 2Qaeyz > Qayt 274 2cyz > > 2zy(1+z)2zu satisfy the condition Et+yte= 2z then sytaz+yz>34+V227414J/y2t1lev2274+1 First We solution: have 2z = z + + z > 2w + z => z( #9)” — 2w — z >3 Because the positive root of the trinomial zt? — 2t — z is 14+VJ1+ 2? Z we get Írom here ⁄~ụ È 14+J1+ 2? Z © zVzw> 1+ VW1+z? Of course, we have two other similar inequalities Then, U+#z+Uz > > #ZVWUz+Uvw#z + zVzÈ 3+VWzˆ+1+vw2+1+Vz?+1, and we have both a proof of the given inequality, and a little improvement of it 38 Solutions Second solution: Another improvement is as follows Start from 1 ep 1 p?gtn let 3.2 2.2 +#“z“ 2.2 2.2.2 + “z4 “U27; which is equivalent to (+ #z + 9z)” > 2z (w + + z) ++22z? = 3( + +2)” Further on, (cy +ez+yz—3)? = (cyt az+yz) —6(ay+az+yz)+9> >3(++z)Ÿ—6(øu + zz + z)+9= 3(z2°+?+z°) +9, so that + + øz +z >> 3+ W3(z2+?+z2) +9 But 3(z2+2+ z2) +9 >Vz2+1+V2+1+Vz?~+1 is a consequence of the Cauchy-Schwarz Inequality and we have a second improve- ment and proof for the desired inequality: > 3+v3(z?2+w2+z23)+9> > 3+Wwz?+1+v¿/?+1+Vz?+1 V rytazt+yz 22 [ Laurentiu Panaitopol | Prove that 1+2? 1+? l++z? 1+z? lI+z+z? l+z+ự ? — ? for any real numbers x,y,z > —1 JBMO, Solution: Let us observe that +ự? < and I+#+z 1+z? ` TIẾT — 2003 >0, so 1+z? 1y} and the similar relations Setting a =1+27,b=1+y?,c=14 prove that a b e 3+b Date 2b+a >1 (1) 2’, it is sufficient to Old for any a,b,¢> Let A= A+4C —-2B and Inequalities 2c-+b, B= 2a+e, B+4A-— b=——^ New c= ae 39 4B—2A 2= ST SỐ —*S, Then C= 2b+a and the inequality (1) is rewritten as C A PB BC HA “+C=~+~+4Í[ +1 Because A, B,C > 0, we have from the AM-GM CoA and —-+L 7A + =1 B + —> c3 Inequality that i and the conclusion follows An alternative solution for (1) is by using the Cauchy-Schwarz Inequality: a b € a? b? c (a+b+c)? + + = + + > > 2c+b 2a+c 2b+a 2ac+ab_ 2ab+cb 2bc+ac — 3(ab + be + ca) 23 Let a,b,c > with a+6+c=1 Show that a +b 0ð tee b+e c+a Fay a+b — Solution: Using the Cauchy-Schwarz Inequality, we find that a+b (Soa? +1) b+e ~ SP(b+o+ Soa? + So ab And so it is enough to prove that + `ab 281 wy +2)+ab (Se) > 2S a(0)b The last inequality can be transformed as follows +} > 29a (b+c)+2SÖab @ 1+ IV 1+ +25 ab = +) (Se) and it is true, because » and a? > »3 (Chebyshev’s (S2) )> S” Inequality) > +20" > Soa’, 40 Solutions 24 Let a,b,c > such that at + 64 + ct < 2(a?b? + bc? + c?a”) Prove that a? +b? +c? < 2(ab+ be + ca) Kvant, 1988 Solution: The condition So at < 2S `a?0? 1s equivalent to (a+b+c)(at+b—c)(b+c—a)(c+a—6b)>0 In any of the casesa = 6+ c, b=c+a, Soa is clear So, supposea+bf#¢c, numbers 6+ c—a, c+a—b, c=a+b, the inequality < 2S ab b+cH#a, a+b-—c c+aF b Because at most one of the is negative and their product is nonnegative, all of them are positive Thus, we may assume that a* ————— S22 (i —m) So * „(8i — #rm) But 52 Đ (am > (n— 1) — =4n—-Ñ+— >á4n-— (r-=†tz=mr=m)È n n a a — a1) i=1 so 5+ So (ai — #m) = = »- — #¿11 — mod i=1 nm » —#¿+1)Ÿ > 4n — 21 and the problem is solved 32 [ Murray Klamkin ] Find the maximum value of the expression x? x2 + x3%3 + + + a? jay +0722, when 21, 2%2, ,%n-1,%n > add up to and n > Crux Solution: First of all, it is clear that the maximum „#2 — c;3 wir attained for 7; = Mathematicorum is at least 2m because this value 1s — -'-—#„ = Now, we will prove by induction that 2 %12 + 45 + + !+ T đ 1u T81 2x Old for all 41,2%2, ,%n—1,2%n and New Inequalities 47 > O which add up to Let us prove first the inductive step Suppose the inequality is true for n and we will prove it for n + We can of course assume that x2 = min{2%1,2%2, ,£%n41} But this implies that 2 Ujtg + egag t+ +a7a 2 on + U7,2 (21 + 22) < (a1 +22)" 43 + 45044 °+°4+ 05 But from the inductive hypothesis we have (đi +22) #3 + 2314 + + +12 18a + #2 (#1 + #a) < > and the inductive step is proved Thus, it remains to prove that a2b + bŸc + ca < ifa+b+c=1 We may of course assume that a is the greatest among a, b,c In this case the inequality ab + b2e + ca abe > 67c,ae > ac < (a + a (0 + 5) follows immediately from Because a +; = 1=—22 a a+ = 2 we have proved that a2b+ b?e+ c?a < and this shows that the maximum is indeed 27 33 Find the maximum value of the constant c such that for any Z1,#2, ,p,''' >0 for which xp41 > 2, + 4% + -+ a, for any k, the inequality V1 + V#s + - + VW#n < €V#1 + øa + : +2 also holds for any n IMO Shortlist, 1986 Solution: First, let us see what happens if zz¿¡ and #1 -Ƒ#a +- +#¿ are close for any k For example, we can take x, = 2", because in this case we have x1 +ao+: +a,% = ©p41—2 Thus, we find that n c> › v2 k=1 for any n Taking the limit, we find that c > 1+ V2 Now, let us prove that 1+ /2 works We will prove the inequality V?t + V#s + - + Vin < (14+ V2)Vai Fao + Fon 48 Solutions by induction For n = or n = it is clear Suppose prove that v/#1 + 4⁄22 + : +4/#„ T /#a+i < (1 + it is true for n and we will ⁄2)⁄21 +#a + ': +#np T#an+1 Of course, it is enough to prove that Vtnt+1 < (14+ v2) (2i + #2 T-'* †#n‡L— V41 +#a +: C#n) which 1s equivalent to V?\ +8 + +** 8n + Vi +8 + tem < (14 V2) Vent But this one follows because #1 ta T - Ð+đa € #n+1 34 Given are positive real numbers a,b,c and x,y,z, for which a+ a2 =b+y= c+2z=1 Prove that 1 (abe + xyz) (— ay +—+ —) bz > cx Russia, 2002 Solution: Let us observe that abc + xyz = (1 — 6)(1—c) + ac+ab—a l-e c a abc + xyz pn T 1—b Thus, a(1 — b) Using these identities we deduce immediately that 3+ (rye +abe) (“+ b+ 2) = ay bz c@ a, l-c l-a Now, all we have to is apply the AM-GM a l-c + b l-a + Cc 1-6 ¢ 1-6 toe tra a b tn) Cc Inequality lve, a lea b TP c Ộ 35 [ Viorel Vajaitu, Alexandru Zaharescu | Let a,b,c be positive real numbers Prove that ab + s+b+2c First be b+ec+2a + ca < Ì(a+b+e) c+a+92b—4 Gazeta ‘ Matematica solution: We have the following chain of inequalities ab tba ab Tae ab SLT 1 (te) at+b+e Old Second and New Inequalities 49 solution: Because the inequality is homogeneous we can consider without loss of generality that a+ 6+c¢=1 and so the inequality is equivalent to 1 a(œ+ 1) — 4abe We have =—_— t(t+1) t+1 t , So the inequality is equivalent to 2a —< < 1 anit tube We will prove now the following intercalation: 2a -9 and the identity (a + 1) =4 The inequality in the left can be written as » 1+ 9abe , which is exactly Schur’s Inequality ab< 36 Find the maximum value of the expression a”(b+ec+đd) +~b?(e+d+a) +c(d+a+b) + dđ”(a+b+ e) where a,b,c,d are real numbers whose sum of squares is Solution: The idea is to observe that a?(6+ce+d)+6?(c+d+a)+c?(d+a+b)+d?(at+b+c) is equal to » ab(a” +b*) Now, because the expression ab(a? +6?) appears when writing (a — b)*, let us see how the initial expression can be written: Sabla ba?+?) + 6°) = » SoS 44 54 +4 6a7b? — (a — b)* 3À 2a2+6S a??— À The maximum is attained fora =b=c=d= = (a -b)* — 3—š(a—b) ˆ