Uiversity of Maitoba, Mathletics 009 Sessio 5: Iequalities Facts ad defiitios AM-GM iequality: For a, a,, a 0, a + a + + a (a a a ) /, with equality iff all a i s are equal Cauchy s iequality: For reals a,, a, b,, b, if u = (a,, a ) ad v = (b,, b ), the u v u v, or equivaletly (a b + + a b ) (a + + a )(b + + b ), where equality holds iff (a,, a ) ad (b,, b ) are proportioal Covex fuctios, Jese s iequality: For x,, x I, where I R is a iterval, if f is a cotiuous fuctio o I which is covex (cocave up), the ( x + + x ) f f(x ) + + f(x ) If f is strictly covex, the equality holds iff all x i s are equal If the fuctio is cocave (cocave dow), the the sig of the iequality is reversed If the fuctio is twice differetiable, it is covex iff f 0 o the iterval I practice, it is usually easier to show that f is mootoe icreasig The sum of two covex (cocave) fuctios is a covex (cocave) fuctio Multiplicatio by a positive umber also preserves covexity (cocavity) Weighted Jese s iequality: If x,, x I, where I is a iterval, λ,, λ > 0, λ + + λ =, f is covex o I, the λ f(x ) + + λ f(x ) f(λ x + + λ x ) If f is strictly covex, the equality holds iff all x i s are equal Corollary: weighted AM-GM iequality: If x,, x 0, λ,, λ > 0, λ + +λ =, the λ x + + λ x x λ xλ, equality holds iff x = x = = x Power mea iequality: Let x,, x 0, λ,, λ > 0, λ + + λ = For t R, t 0, defie the weighted mea M t of order t as M t := (λ x t + + λ x t ) /t Also M 0 := x λ x λ = lim t 0 M t, M := mi{x,, x } = lim t M t, M := max{x,, x } = lim t M t The M s M t, if s < t
Examples Example 5: For a, b, c 0, prove a 3 + b 3 + c 3 a b + b c + c a Solutio: AM-GM iequality implies a 3 + b 3 = a 3 + a 3 + b 3 3 3 a 6 b 3 = 3a b Similarly, we obtai b 3 + c 3 3b c ad c 3 + a 3 3c a Addig all three iequalities, we get the required oe Example 5: For each positive iteger, fid the smallest possible value of the expressio (x + + x )(x + + x ), over all possible x,, x > 0 Solutio: With x = x = = x =, our expressio is equal to So, it is eough to prove that (x + +x )(x + +x ) for ay x,, x > 0 This follows immediately from the Cauchy s iequality with a k = x k ad b k = / x k, k =,, Example 53: If a, b 0, a + b =, prove ( + 5 a) 5 + ( + 5 b) 5 64 Solutio: The fuctio f(x) = (+ 5 x) 5 is cocave o [0, ) as f (x) = 5(+ 5 x) 4 5 x 4/5 = (x /5 + ) 4 is decreasig there By Jese s iequality, ( + 5 a) 5 + ( + 5 b) 5 = f(a) + f(b) f() = 64 Example 54: For a, b, c > 0, prove a0 + b 0 + c 0 ( ) a + b + c 5 a 5 + b 5 + c 5 3 Solutio: (Hit: rewrite usig M 0, M 5 ad M ) Let M t be the mea of order t of a, b, c, so a t + b t + c t = 3M t t Our iequality becomes 3M 0 0 3M5 5 M 5, which is equivalet to M 0 0 M 5 M 5 5 To obtai this iequality we multiply the fifth powers of the power mea iequalities M 0 M ad M 0 M 5
3 Problems for discussio Discussio problem 5: For a,, a 0, prove a 5 + + a 5 a 3 a a 3 + a 3 a 3 a 4 + + a 3 a a Discussio problem 5: For x,, x > 0, prove x x + x + x x + x 3 + + Discussio problem 53: If a, b, c > 0, prove Discussio problem 54: If a,, a, prove x x + + x x + x a a + 3b + 3c + k= + a k b 3a + b + 3c + Discussio problem 55: For a,, a > 0 with a a =, prove a + a + + a + + a a c 3a + 3b + c 3 7 3
4 Solutios for discussio problems Discussio problem 5: For a,, a 0, prove a 5 + + a 5 a 3 a a 3 + a 3 a 3 a 4 + + a 3 a a Solutio: Let a + = a ad a + = a AM-GM iequality gives 3a 5 j + a5 j+ + a5 j+ a 5 5 j 5 a5 j+ a5 j+ = a3 ja j+ a j+, j =,, Addig the above iequalities, we obtai the required oe Discussio problem 5: For x,, x > 0, prove x x + x + x x + x 3 + + x x + + x x + x Solutio: Let x + = x Applyig Cauchy s iequality with a k = xk + x k+, k =,,, we get x k xk + x k+, b k = ( x + + x )( ) (x + x ) + (x + x 3 ) + + (x + x ) (x + + x ), x + x x + x which becomes the iequality that we eed to prove after divisio by (x + + x ) a Discussio problem 53: If a, b, c > 0, prove a + 3b + 3c + b 3a + b + 3c + c 3a + 3b + c 3 7 Solutio: Fix S = a + b + c Take f(x) = x/(3s x) The iequality becomes f(a) + f(b) + f(c) 3f(S/3), ie, Jese s iequality It remais to show that f is covex o [0, S) (as a, b, c > 0, each of them is smaller tha S) Ideed, it is clear that f (x) = is icreasig o [0, S) (eve o [0, 3 S)) Discussio problem 54: If a,, a, prove k= + a k + a a 3S (3S x) Solutio: Cosider f(x) = /(+e x ) Straightforward computatio shows that f = ex (e x ) ( + e x ) 3 > 0 for x 0, so that f is covex o [0, ) Sice l a k 0, k =,,, by Jese s iequality k= + a k = f(l a k ) f((l a + + l a )/) = k= Discussio problem 55: For a,, a > 0 with a a =, prove a + a + + a + k= + a a Solutio: We write k ak a k = k k, k =,,, ad use the weighted AM-GM iequality: kk ak ( + ) ( a k k a ) (+) kk 3 3 = + ( 3 ) (+) 3 3 + 4
5 Take home problems ( a + b + c ) a+b+c Take home problem 5: If a, b, c > 0, prove a a b b c c 3 Take home problem 5: For a,, a, b,, b 0, prove ((a + b ) (a + b )) (a a ) + (b b ) Take home problem 53: For a,, a > 0, prove a + + +a + a a (a + +a ) 5
6 Take home solutios Take home problem 5 solutio: As all the umbers are positive, we ca take the logarithm o both sides to arrive at the equivalet iequality a l a + b l b + c l c (a + b + c) l((a + b + c)/3) = 3((a + b + c)/3) l((a + b + c)/3), which is Jese s iequality with f(x) = x l x for a, b, c, if f is covex o (0, ) Ideed, f (x) = l x is icreasig o (0, ) Take home problem 5 solutio: If for some k we have a k = 0, the the iequality becomes obvious Hece, we ca assume that all a k are strictly positive, k =,, Dividig our iequality by (a a ), we obtai a equivalet iequality (( + b ) ( + b )) a a ( b + b ) a a Take the logarithm o both sides ad deote y k = b k /a k 0, k =,, Now we eed to prove (l( + y ) + + l( + y )) l( + y y ) The fuctio f(x) = l( + e x ) is covex o [0, ) because f (x) = +e x is icreasig o [0, ) Hece, by Jese s iequality l( + y ) + + l( + y ) = f(l y ) + + f(l y ) f(l((y y ) )) = l(+ y y ) Take home problem 53 solutio: Let M t be the power mea of order t for a,, a Our iequality ca be rewritte as M+ + M 0 M, which follows immediately from the power mea iequalities M + M 0 (ad so M+ M 0 ) ad M + M Alterative way: AM-GM iequality implies a + k + + a + j + a k j= a j, for k =,, Addig these iequalities, we obtai the required oe j= 6