this chapter, we introduce some of the most basic techniques for proving inequalities. Naturally, we have two ways to compare two quantities:

Transcription

1 Chapter 1 Proving lnequal ities ~~_,. Basic Techniques for _ Among quantities observed in real life, equal relationship is local and relative while unequal relationship is universal and absolute. The exact nature of inequality is the studying of unequal relationship among quantities. Now, for two quantities we can always compare them, or try to show one of them is greater than the other: in other words, we need to prove an inequality. The techniques for proving an inequality varies from case to case and often require some basic inequalities such as the famous AM-GM Inequality and the Cauchy-Schwarz Inequality; other techniques even involve some more advanced algebraic rearrangements. In this chapter, we introduce some of the most basic techniques for proving inequalities Direct Comparison Naturally, we have two ways to compare two quantities: (1) Compare by subtraction: to show A ~ B, it suffices to show A -B ~0; (2) Compare by division: say B > 0, to show A ~ B, it suffices to A show B ~ 1. When we use the above two methods to compare two quantities, usually some forms of rearrangements is required. For example, factorization, separating and combining terms are some of the most used tricks.

2 2 Methods and Techniques for Proving Inequalities Eg. 1 Let a, b, c be positive real numbers, prove that a 2 +be +b 2 +ca +c 2 +ab >- +b + b+c c+a a+b ~a c. Proof. The LHS-RHS is a 2 +be -a +b 2 +ca-b +c 2 +ab-c b+c c+a a+b = a 2 +be -ab -ac +b 2 +ca -be -ba +c 2 +ab -ca -cb b +c c +a a +b =(a -b)(a -c)+ (b -c)(b -a)+ (c -a)(c -b) b +c c +a a +b (a2 -b2)(a2 -c2) + (b2 -c2)(b2 -a2) + (c2 -a2)(c2 -b2) (b +c)(c +a)(a +b) a4 + b4 + c4 - a2b2 - b2c2 - c2a2 (b +c)(c +a)(a +b) (a2 -b2)2 +(b2 -c2)2 +(c2 -a2)2 = 2(b +c)(c +a)(a +b) ~o. therefore the original inequality holds. Eg. 2 For real numbers x, y 1 z that satisfies xy + yz + zx = - 1~ show that: Proof. Because x 2 +5y 2 +8z 2-4 = x 2 +5y 2 +8z 2 +4(xy + yz +zx) = (x +2y +2z) 2 +(y -2z) 2 ~0. Wehavex 2 +5y 2 +8z 2 ~4. Eg. 3 Let a 1 b 1 c be positive real numbers. Prove that for any real numbers x, y, z, we have: 2,) ab c ( {a+b {b+c {C+a ) ~ (a +b)(b +c)(c +a) 'V~-c-xy +'\j----;--a-yz +'\j---,;--b-zx When does the equality hold?

3 &sic Techniques for Proving Inequalities 3 Hint It is well known that we will use similar method to prove the above problem. Proof. UIS-RHS = lb!cx2 + c ~ay2-2j (b +c~c +a) xy l + [ c 2 + b 2 2 / be ] + [ c + ~y ~z - 'V (c +a)(a +b)yz b +ex 2 a +bz a 2-2 Jcb+c)Ca+b)xzl = ~abljac:+c) -./bc:+a)r ~ 0 (here.2:; represents a cyclic sum). Hence the original inequality holds. Eg. 4 Let a, b, c be positive real numbers. Show that: Proof. Since this inequality is symmetry in a, b, c, WLOO we can assumea ~b ~c, thus: Hence Note. From this problem we obtain: Generally, ifx, E R+, i = 1, 2,, n, we have: the proof is similar to what we have done previously. Eg. 5 Let a, b, c be positive real numbers such that a 2 +b 2 +c 2 = 1. Find the minimum for

4 4 Methods and Techniques for Proving Inequalities Hinl When a = b = e, S = 3. So we guess S > 3. In fact, S -3 = _1 +_1 +_1-3 _2(a3 +b3 +e3) ~ ~ ~ ~e =az +bz +ez +az +bz +cz +az +bz +cz -3-2(a2 +b2 +c2) a 2 b 2 e 2 be ea ~ = a2 (_1 +_l)+b2 (_1 +_1)+ e 2 (_1 + _1) _ 2(a 2 + b 2 + c 2 ) b 2 c 2 a 2 c 2 a 2 b 2 be ca ~ = az ( i -! f + bz (! - ~ f + cz ( ~ - i f > 0. Therefore, the minimum for S is 3. Note. Don't be afraid to guess the right answer (which is quite effective in handling extremum problems)! Beside guessing the right answer, we should also guess when the equality holds and keep that in mind when we later try to prove the inequality Method of Magnifying and Reducing If we get stuck proving an inequality like A :< B directly, we can try to find a quantity C that acts like a bridge: i.e. if we have both A :< C and C :< B, then A :< B naturally follows. In other words, we can magnify from A to C, and magnify again from C to B (the same idea applies to reduction). The trick here is to find a suitable level of magnification or reduction. Eg. 6 Assume n is a positive integer and au azo..., an are positive real numbers. Show that " 1 1 ~ =---~ i = l at (at +a2 + +a,) 2 > 2 n [ i: _ :.. ]. ( n +2011) i=t a 1 a 1 +a 2 + +a,

5 &sic Techniques for Proving Inequalities 5 Proof. From Cauchy's Inequality, Hence It suffices to show that which is equivalent to n (n ) (_!_ + _1_ + + _1_ + 1 a1 az a,. n(at +az + +a,) ~ (n ) [_!_ +_!_ + + _!_ ], at a2 a,. at +a2 + +a,. which is equivalent to ) which is equivalent to ~ (n 3 + 1) ~ _!_ + (2011n ) + ;~1 a, '"' LJa; i=t " 1 1 (2011n - 1) ~ - ~ (2011n - 1)n 2 -,.-, ;~1 a; '"' LJa; ;~t

6 6 Methods and Techniques for Proving Inequalities or so the conclusion holds. n " 1 2 ~a;~- ~n, i=1 i=1 a; Eg. 7 Prove that for all positive real numbers a, b, c, we have: Proof. Sincea 3 +b 3 =(a +b) (a 2 +b 2 -ab) ~(a +b)ab, we have: 1 ~ 1 c a 3 +b 3 +abc """"ab(a +b) +abc abc(a +b +c)" Similarly, 1 ~ a b 3 +c 3 +abc ""'abc(a +b +c)' 1 ~ b c 3 +a 3 +abc ""'abc(a +b +c)" Adding the above three inequalities together, we obtain: Note. When we prove inequality with fractional parts, we seldom change the fractions to a common denominator. Instead, we can simplify the fractions by means of magnifying or reduction. Eg. 8 Assume that a; ~ 1(i = 1, 2,, n). Prove that: Hint. Observing the two sides of the above inequality, how can we manage to get 2" out from the LHS? Proof. C1 +a1)(1 + a z) C1 +an) = 2,. ( 1 + a1; 1) ( 1 + az; 1)... ( 1 +a,.; 1).

7 &sic Techniques for Proving Inequalities 7 Since a; - 1 ~ 0, we have: (1 +a1)(1 +az) (1 +a,.) ~ 2,. ( 1 + a1 ; 1 + a2; a,.; 1) ~ 2,. ( 1 + a1-1 + a a,. -1) n+1 n+1 n+1 2" = n + 1 (1 +a1 +a a,.). Thus the original inequality holds. Eg. 9 Find the maximum real number a, such that x +./yz +z2 Y + z >a holds for all real numbers x, y, z../xz + zz./xz + yz Soln 1. Let x = y, z-0, we find LHS -2. As a result, a >2 will lead to a contradiction. Hence a ~ 2. WLOG, we assume x ~ y ~ z, let us prove that Move Y, z to the RHS, the above inequality is./zz + xz Jxz + y z equivalent to or Omitting x from both sides. Since J x 2 + y 2 + y > 2y, J x 2 + z 2 + z > 2z, we only need to show

8 8 Methods and Techniques for Proving Inequalities ;-=:::;:.x:::::::::;:- increases when x Jxz +zz increases. Similarly, x increases when x increases. Judging by lx2 + Y2 these two facts, we only need to consider the case when x = y. Let x = y, we need to prove The above inequality is equivalent to or which is obvious. Therefore, SoalliiiX = 2. Note. We can also use undetermined coefficients to solve this problem. Soln 2. Again, we show that Assume

9 &sic Techniques for Proving Inequalities 9 where a is an undetermined coefficient. Note that (J) is equivalent to (xa +ya +z.. ) 2 ~4x2a- 2 (y 2 +z 2 ). Since (x.. + y.. + z.. ) 2 ~ 4x"' (y"' + z"'), we only need to guarantee It's easy to see that a = 2 fits the bill. Thus, Hence, a"'"" = 2. Eg. 10 Assume non-negative real numbers a1, a2,..., a,. and b1, b2,..., b,. both satisfy the following conditions: " (1) :L:Ca; +b;) = 1; i=1 " (2) :L:i(a; -b;) = 0; i=1 n (3) 2::: i 2 (a; + b;) = 10. i=1 10 Show that for any 1 ~ k ~ n, max{ak, bk} ~ 10 + k2' Proof. For any 1 ~ k ~ n, we have ~ (10 -k 2 ak) (1 -ak) = 10 - (10 +k 2 )ak +Pai, hence a1 ~ 10 ~ p. Similarly b1 ~ 10 ~ k 2, thus

10 10 Methods and Techniques for Proving Inequalities Eg. 11 Positive real numbers x, y, z satisfy xyz > 1, show that (2015 IMO) Proof. The original inequality is equivalent to have From Cauchy Inequality and the given condition that xyz > 1, we or x2 + Y2 + z2 yz + Y2 + z2 xs + y2 + z2 ~ x2 + y2 + z2 Similarly, Adding these three inequalities together and use the fact that x 2 + y 2 + z 2 > xy + yz + zx, we have Comment. Contestant Boreico Iurie from Moldova has won the special price for his solution outlined below. Since As a result, x2 (x3-1)2 (y2 + z 2) x3(xs + Y2 +z2)(x2 +y2 +z2) >O.

11 &sic Techniques for Proving Inequalities 11 = xz + :z + zz ~ ( x 2 - ; ) ~ 2 + \ + 2 ~(x 2 -yz)(sincexyz ~1) X y Z eye ~ Analyzing the inequality To analyze an inequality t we first assume that it holds and deduce from it a series of equivalent inequalities (i.e. we require that each step is reversible) until we reach an inequality that is easier or more obvious to prove than the original. Such method is usually quite helpful in gathering proving thoughts. Eg. 12 If x, y E R, y ~ 0, andy (y + 1) ~ (x + 1) 2. Prove that Proof. If 0 ~ y ~ 1 t Now t y(y -1) ~x 2 then y (y -1) ~ 0 ~ x 2 if y > 1, from the assumptions we have y(y +1) ~ (x +1)2, y ~ J (x + 1) 2 + ~ - ~. To prove y (y - 1) ~ x 2, it suffices to show that holds. Jcx +1)2 + ~ - ~ ~Jxz +~ +f, ~ (x + 2 1) + ~ ~ x 2 + ~ + 2J x 2 + ~ + 1 t ~ 2x ~ 2J x 2 +!. The final inequality is obvious, hence the original inequality Eg. 13 Assume a, b, c E R+. Show that a +b +c -3~ ~a +b -2../(ib.

12 12 Methods and Techniques for Proving Inequalities Proof. Note that a +b +c -3~ ~a +b -2.;;;Ji Because c + 2.;;;Ji = c +.;;;Ji +.;;;Ji ~ 3:; c,;;ib,;;ib = 3 ~. Therefore, a +b +c -3~ ~a +b -2.;;;Ji. Note. To prove an inequality, sometimes we need to alter between the analyzing method and the comprehensive method. For Eg. 13, we see from the analyzing method that c + 2.;;;Ji ~ 3 ~ (should be true). If we insist on continuing with the analyzing method, we might produce more complications. On the other hand, the comprehensive method leads us the solution. Eg. 14 Assume n E N+. Prove that _1_(1 +_l )~1_(_1_ +_l ). <D n n - 1 n 2 4 2n Proof. To prove <D, it suffices to show n(1 +1_ ) ~ (n +t)(_l +_l + +!_). 3 2n n The left hand side of is The right hand side of is -+-+n n n ( ) n -1 n(1_ +1_ +.. +!_)+ (_l +1_ +.. +!_) 2 4 2n 2 4 2n = I!:.. + n (1_ !_) + (1_ + 1_ !_) n 2 4 2n

13 &sic Techniques for Proving Inequalities 13 Compare if n ~ n' and oo +-- ::>-:- +oo n - 1,_ 4 2n are both true (which is quite obvious), then holds. In conclusion, the original inequality CD is true. Eg. 15 Let a, b, c be positive real numbers such that abc = 1. Prove that (a +b)(b +c)(c +a) ~ 4(a +b +c -1). Hint The idea is to treat a as a parameter and regard the inequality as a quadratic equation, then we can use the discriminant of the quadratic equation to prove the inequality. Proof. WLOG, assume a ~1. The original inequality is equivalent to that is, (a 2-1)(b +c) +b 2 (c +a) +c 2 (a +b) +6 ~4a +3(b +c). Since (a+ 1) (b +c) ~ 2.[a 2../bC = 4, if we can show 4(a -1) +b 2 (c +a) +c 2 (a +b) +6 ~4a +3(b +c), then CD holds. Now, is equivalent to 2 +a(b 2 +c 2 ) +bc(b +c) -3(b +c) ~0, so we only need to show that ~ (b +c) 2 +(be -3)(b + c) +2 ~0. Define f(x) = ~ x 2 +(be-3)x + 2, then its discriminant

14 14 Methods and Techniques for Proving Inequalities or 6. =(be -3) 2-4a. It suffices to show 6. ~ 0, which is equivalent to (~ -3f -4a ~o. or (a -1) 2 (4a -1)? 0. Since a? 1, the above inequality is obviously true, hence Q) follows. From the above discussion, we see that the equality holds when a = b = c = Method of Undetermined Coefficients Many times, we can also introduce undetermined coefficients and solve for these coefficients to prove the inequality. Eg. 16 Assume x, y, z are real numbers that are not all 0. Find. xy +2yz the maxrmum value for X y Z H ml. T o f" m d h. f xy + t e maxrmum or 2 yz 1 d , we on y nee to X y Z show that there exists a constant c, such that xy +2yz ~ 2+2+2"""c' X y Z CD and that the equal sign holds for some x, y, z. CD can be translated to x 2 + y 2 + z 2?!_ (xy + 2yz). Since the c right hand side has terms xy and 2yz, we split the term y 2 in the left hand side intoay 2 and (1 -a)y 2 Since,

15 &sic Techniques for Proving Inequalities 15 We want 2 ~ = 2, which givesa = ~. 2ra So ln. Since, We obtain or The equality holds when x = 1, y =./5, z = 2, hence the maximwn 1 can be reached. Eg. 17 For ; ~ x ~ 1, find the maximum value of Soln. Let us consider the maximum value of [a (1 + x) ] 5 [BO - X) J [ r ( 2x -1) ] 2 ' where a ' /3' r are positive integers satisfying Sa - /3 + 4y = 0, ac1 +x) =po-x) = y(2x -1). This implies ~ _Ji.±L f3 +a - 2y +fi' Plugging in f3 = Sa + 4y, we have 0 = 2(3ay +Sa 2-2y 2 ) = 2(Sa -2y)(a +y), Let (a, p, y ) = (2, 30, S), from AM-GM Inequality, we obtain

16 16 Methods and Techniques for Proving Inequalities [2(1 +x)] 5 [30(1-x)][5(2x -1)] 2 ~(~f. The equality is achieved when x = ~. As a result, the maximum y. ss value for (1 +x) 5 (1-x)(1-2x) 2 is~. Eg. 18 C Ostrowski) Assume two sets of real numbers a 1, a2,, a,. and b 1, b 2,, b,. are not scaled version of each other. Real numbers x 1, x2,, x,. satisfy: Prove that a1x1 +a2x2 + +a,.x,. = 0, b1x1 +b2x2 + +b,.x,. = 1. xf +X~ + "' +X~ ~,_i - --' ~af ~bt- (~a;b;) 2 i-1 i-1 i-1 " " " " Proof 1. Assume ~xr ~xt +a ~a;x; +,B(~b;x; -1), i-1 i - 1 i - 1 i-1 where a,.b are undetermined coefficients. Thus, ~xr = ~ (x; + aa; +fib; ) 2 _ ~ Caa; +,Bb;) 2 _.B i-1 i-1 2 i-1 4 For the above inequality, the equal sign holds if and only if x; =- ' aa +,Bb ' (i = 1, 2,..., n). 2 n " Substitute CD back into ~ a;x; = 0 and ~ b;x; = 1, we have i-1

17 &sic Techniques for Proving Inequalities 17 n n n Here, A =~at, B = ~bt, C = ~a,b,. Therefore, i=t i=t i=t Hence 2C 2A a= AB -C2 ' {1 =-AB -c2 Note 1. There are two more ways to prove the inequality, for reference, we will mention them as follows: Proof 2. From the Cauchy-Schwarz Inequality, for every t E R, we have [_t (a;t +b;) 2 J. (xr +x~ +. +x~) ~ [ t (a;t +b;)x, r t=l Or t=l (xi + x~ + + x: )(At 2 + 2Ct +B) - 1 ~ 0 = 1. always holds true. From tl. ~ 0 (tl. is the discriminant with respect tot ), we obtain Proof 3. (a combination of the previous two proofs) According to the conditions, for any). E R, we have " ~ (b; - ).a 1 )x 1 = 1. i=t By Cauchy-Schwarz Inequality, " " [ " ]2 ~xt ~ (b, - ).a 1 ) 2 ~ ~ (b, - ).a 1 )x 1 = 1. i=t i=t i=1 n 2 1 Hence, ~x, ~ B + 2A -2).C -1 ). Recall that our goal is to prove

18 18 Methods and Techniques for Proving Inequalities So we only require that " 1 ~ xr ~ ::.--,c""'2:-. i=t B A or A 2 A 2-2ii.AC + C 2 ~ 0. Picking A = ~ satisfies the above requirement. Note 2. From this problem one can show the Fan-Todd Theorem: Let a.~o, b.~o (k = 1, 2,..., n) be two sets of real numbers not scaled of each other. Also assume that a ;b k =I= a kb; (i "=I= k), then To prove the Fan-Todd Theorem, it suffices to set x.~o = (~ )- 1 ~ b ar b, We ask the reader to Check that Xt, X2, "', X,. ~k ar k - a.~o r satisfies the conditions. Eg.19 Find the maximum m,. for the function Express m, using m and find limm,...-oo Hint. Every denominator inf, is rather complicated. To proceed, we should first simplify the denominators using substitutions. Soln. Let a; = 1 + _; +, 1 ~ i ~ n. Define a,.f-1 = 1. Xi Xn

19 &sic Techniques for Proving Inequalities 19 Thus, 1 +x, +x,+t + +x,. 1 a; Also, 1 +x;+t +x; x,. 1 Hence X.=--- a; a;+t Substituting x, 's, we have Th find the maximwn for f,., we construct the following inequalities: 2 a 1 + J.ta2 ~ 2A1 a 1, a2 a~ 2 - +,ha3 ~ V..2a2, a3 2 ~n +A~ ~ 2\,.a,. Here At, A2,, A,. are parameters, A; ~ 0. Adding CD, if we let 2).1 = 1, ~~--~-1 +J.t' 1 v.... = 1 + A;-1' then immediately j,. < ;.:. Note that;., ~ A;-1, ando <A; < 1. Therefore, lim;.,. exists. It's,.-oo

20 20 Methods and Techniques for Proving Inequalities easy to see that the limit is Normalization When the inequality has same order in its terms, we can assume that the variables add up to a constant k. In doing so, we simplify the inequality and at the same time enhance the known conditions, both of which help us to solve the problem. Eg. 20 Assume a, b, c are positive real numbers, prove that Proof. Since each term on the left hand side is of the same order, WLOG we assume a + b + c = 3. It suffices to show (a +3) 2 (b +3) 2 (c +3) 2 ~ 2a 2 + (3 -a) 2 + 2b 2 + (3 -b) 2 + 2c 2 + (3 -c) 2 """" 8 Define Then /( ) (x +3)2 E a+. X = 2x2 + (3 -X )2 ' X /( ) x 2 +6x +9 x = 3(x 2-2x +3) 1( 8x+6) 1( 8x+6) = x 2-2x + 3 = (x -1) ~ ~ ( 1 + 8x t 6) = j ( 4x + 4). So j(a) +/(b)+ / (c)~ j (4a +4 +4b +4 +4c +4) = 8. Eg. 21 Assume a + b + c > 0, and ax 2 + bx + c = 0 has positive real number root. Prove that

21 &sic Techniques for Proving Inequalities 21 4min{a, b, c} <a +b +c <: max{a, b, c}. Proof. WLOG assume a + b + c = 1, otherwise we can replace a, b, c by a+~ +c, a+: +c, a+~ +c respectively. First, let us prove the statement 4 max{a, b, c} >9 4 (1) If b > 9, then the statement already holds true. (2) If b <!, since b 2 > 4ac, we have ac < 8 ~. But a + c = 1 - b > ~, so if a < 0 or c < 0, we will have c > ~ or a > ~, the statement is thus true. If a, c > 0, then ( c ) c < ac < Hence, c < 9 or c > 9. If c < ~, then a >!, the statement is true. Next we prove the statement min{a, b, c} <!. (1) If a "<!, the statement is already true. 1 3 (2) If a > 4, then b 2 > 4ac > c, b + c = 1 - a < 4. WWG assume c > 0 (otherwise the statement is obviously true). Sorc + c > b + c <!, or ( rc +! ) (rc - ~ ) < 0. So c <!, the statement is true. Note. The bound we present is the best: quadratic equation! x 2 + ~ x +! = 0 demonstrates that : cannot be any smaller, while

22 22 Methods and Techniques for Proving Inequalities equation! x 2 + ; x +! shows that 4 cannot be any larger. Exercise 1 1. Let x t y t z be real numbers. Prove that (x2 + y2 + z2)[(x2 + y2 + z2)2 _ (xy + yz + zx)2] ~ (x + y +z) 2 [(x 2 + y 2 +z 2 )- (xy + yz +zx)] 2 2. Let m t n be positive integers and m > n t prove that 3. Assume a, b, n are positive integers bigger than 1. A.. -t and A,. area-ary, B,.-1 andb,. are b-ary. A..--1, A,., B..--1, B,. are defined as: A,. = x,.x,...-1"'xo t A = x..-1xn-2'"xo (written in a-ary form) B,. = x,.x,.-t.. 'Xo, B.. -t = x.. -tx 'Xo(written in b-ary form) where x,. =I= 0, x..--1 =I= 0. Prove that when a > b, we have A,.-1 < B..--1 A, B,.. 4. Let a, b, c be positive real numbers, prove that a 8 + b 8 + c b +- ~ 3b3 3 a c a c 5. Assume real numbers a1 t a 2, t a 100 satisfy: (1) a1 ~a2 ~... ~a10o ~ 0; (2) a1 +a2 ~ 100; (3) a3 +a atoo -< 100. Find the maximum forar +a~+ +aroo. 6. Assume 5n real numbers r;, s;, t;, u;, v; are all bigger than 1 1 " 1 n 1 " (1 ~t ~n). DenoteR =- ~ru S =- ~su T =- ~tu n 1=1 n =t n i =t

23 &sic Techniques for Proving Inequalities 23 U = _!_ ~ u;, V = _!_ ~ v;. Prove the following inequality: n i=t n i=1 ir T;S;t;U;V; ~ 1 ;? (RSTUV ~ 1 )". i=1 T;S;t;U;V; 1 RSTUV 1 7o Assume k, n are positive integers, 1 < k <no x 1, x 2, ooo, xk are k positive real numbers whose sum equals their product. Show that x~- + 1 x~ o If a, b, c E R, show that xr 1 ~ kn. 9o Prove that for any c > 0, there exists positive integer n and a complex sequence a 1, a 2,, a,, such that 1Etat+E2a E,a, 1<(~ Ia; lt)t j=! wheree; E {-1, 1}, j = 1, 2, ooo, n. 10o Assume a, bare positive,() E ( 0, ; )0 Find the maximum for y = a v' sin () + b v' cos (). llo Assume there are n real numbers, each of them has an absolute value not greater than 2 and their cubes add up to 0. Show that their sum is not greater than ; n. 12o Assume that n ~ 3 is an integer, real numbers Xt, x2, " x, E [ -1, 1] satisfy the relationship ~ x~ = Oo Prove that 4=1 13o Assume that the arithmetic average of n real numbers x 1, x 2, x, isa. Show that

24 24 Methods and Techniques for Proving Inequalities 14. Let x, y, z be non-negative real numbers. prove that x(y +z -x) 2 +y(z +x -y) 2 +z(x +y -z) 2 ~3xyz. Point out when the equality holds. 15. Assumen ~2 is an integer, real numbersa1 ~a2 ~ ~a,.> 0, bl ~ b2 ~ ~ b,. > 0. Also assume a1a2"'a,. = btb2"'b,.,.l; (a; -a;) ~.L; (b, - t..;;i<;"" t..;;i<;"" b; ). Show that L; a; ~ (n -1).L; b;. i~l 16. Let x, y, z be positive real numbers. Prove that J 9 (xy +yz +zx) [ (x +y)2 + (y +z)2 + (z +x)2 ~4' i~l