Some Inequalities variations on a common theme Lecture I, UL 2007


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1 Some Inequliies vriions on common heme Lecure I, UL 2007 Finbrr Hollnd, Deprmen of Mhemics, Universiy College Cork, July 2, 2007 Three Problems Problem Assume i, b i, c i, i =, 2, 3 re rel numbers Prove h ( + b + c ) 2 + ( 2 + b 2 + c 2 ) 2 + ( 3 + b 3 + c 3 ) b 2 + c b c b c 2 3 Problem 2 Assume i, b i, c i, i =, 2, 3 re posiive numbers Prove h 3 b c b 2 c b 3 c 3 3 ( )(b + b 2 + b 3 )(c + c 2 + c 3 ), b = 2 b 2 = 3 b 3 Problem 3 Assume i, b i, c i, i =, 2, 3 re posiive numbers Prove h + b + + c 2 + b c b 3 + c 3, b +b 2 +b 3 + c +c 2 +c 3 b = 2 b 2 = 3 b 3 2 A bsic inequliy Theorem Suppose, b > 0 Then 2 b + b, > 0,
2 In oher words, 2 b = min{ + b = b : > 0} = min{s + b : s, > 0, s = } Proof + b 2 b = 2 2 b + b We deduce some consequences Theorem 2 If, b, c, d > 0 Then b + cd ( + c)(b + d), Proof For ll > 0, Hence c = b d 2[ b + cd] + b + c + d = 2[ b + cd] min{( + c) + We cn build on his in wo wys b d b + d = c = + c (b + d) = ( b) 2 (b + d) = ( + c) +, : > 0} = 2 ( + c)(b + d) Theorem 3 Suppose, b, c, d, e, f, g, h > 0 Then 4 bcd + 4 efgh 4 ( + e)(b + f)(c + g)(d + h), e = b f = c g = d h Proof Pu x = b, y = cd, u = ef, v = gh Then 4 bcd + 4 efgh = xy + uv We cn now deduce he following semen (x + u)(y + v) = ( b + ef)( cd + gh) ( + e)(b + f) c + g)(d + h) = 4 ( + e)(b + f)(c + g)(d + h) 2
3 Corollry If, b, c, e, f, g > 0, hen 3 bc + 3 efg 3 ( + e)(b + f)(c + g), e = b f = c g Proof Le d = 3 bc, h = 3 efg nd pply he heorem d + h = 4 bcd + 4 efgh 4 ( + e)(b + f)(c + g)(d + h), whence 3 bc + 3 efg = d + h 3 ( + e)(b + f)(c + g) A soluion of Problem 2 follows Exercise Prove he ls semen Exercise 2 Suppose i, b i > 0, i =, 2,, n Prove h n 2 n + n b b 2 b n n ( + b )( 2 + b 2 ) ( n + b n ), b = 2 b 2 = = n b n 3 The CuchySchwrz inequliy Theorem 2 is specil cse of disguised version of one form of perhps he mos useful inequliy in ll of Mhemics, nmely he CuchySchwrz inequliy Theorem 4 (CuchySchwrz) If x i, y i > 0, i =, 2,, n, hen n n n xi y i ( x i )( y i ), i= i= i= x y = x 2 y 2 = = x n y n Proof Invoke Theorem 2 nd use inducion We cn phrse his differenly Corollry 2 If i, i =, 2,, n re rel numbers, hen n n n 2 i = mx{ i x i : x 2 i = } i= i= i= 3
4 Proof For, if n i= x2 i =, hen n i x i nd here is equliy when i= = = n i x i i= n (i ) 2 (x i ) 2 i= n n ( 2 i )( x 2 i ) i= i= n 2 i, i= x i = i n, i =, 2,, n i= 2 i Theorem 5 If, b, c, d re rel numbers, hen ( + b)2 + (c + d) c 2 + b 2 + d 2 Proof Suppose x 2 + y 2 = Then Hence x + cy 2 + c 2, bx + dy b 2 + d 2 ( + b)x + (c + d)y x + cy + bx + dy 2 + c 2 + b 2 + d 2 I follows h ( + b)2 + (c + d) 2 = mx{ (+b)x+(c+d)y : x 2 +y 2 = } 2 + c 2 + b 2 + d 2 This is specil cse of Minkowski s inequliy Theorem 6 (Minkowski) Assume i, b i, i =, 2, 3 re rel numbers Then n ( i + b i ) 2 n 2 i + n b 2 i i= i= i= Proof Suppose n i= x2 i = Then n 2 i + n n n b 2 i i x i + b i x i i= i= 4 i= i= n ( i + b i )x i i=
5 Hence n n ( i + b i ) 2 = mx{ ( i + b i )x i : i= i= n x 2 i = } n 2 i + n b 2 i i= i= i= 4 Discussion of Theorem 2 The produc b represens he re of recngle A wih dimensions, b, nd cd is he re of recngle B wih dimensions c, d And + c, b + d re he dimensions of recngle C he sum of A, B Thus re(a) + re(b) re(c), A, B, C re scled versions of ech oher Exercise 3 Give geomeric inerpreion of Problem 2 We cn cs he previous semen in differen wy Wih ech poin (vecor) x = (x, x 2 ) is he firs qudrn of R 2, le Then, if y = (y, y 2 ), g 2 (x) = x x 2 g 2 (x) + g 2 (y) g 2 (x + y) Thus g 2 is superddiive on R 2 Noice, oo, h, for every sclr λ > 0, λx = (λx, λx 2 ) nd g 2 (λx) = λg 2 (x), x R 2, ie, g 2 is homogeneous of degree Hence, g 2 is concve on R 2 Anlyiclly, his mens h x, y R 2, nd 0 λ, hen λg 2 (x) + ( λ)g 2 (y) g 2 (λx + ( λ)y) To ge n ide of wh i mens, geomericlly, ry nd picure he surfce in 3spce whose equion is z = xy, x, y 0 Exercise 4 Apropos Problem 2, drw similr conclusion bou he funcion g 3 (x) = 3 x x 2 x 3, x = (x, x 2, x 3 ) R 3, x i 0, i =, 2, 3 In fc, if x = (x, x 2,, x n ) is vecor in nspce nd is coordines re posiive, nd g n (x) = n x x 2 x n, hen g n is superddiive nd is homogeneous of degree on Hence, g n is concve on his region {x R n : x i 0, i =, 2,, n} 5
6 5 Discussion of Theorem 5 I s fmilir h 2 + b 2 is he eucliden disnce from he origin o he poin (, b) in R 2 Thus Theorem 5 is simply n nlyicl semen equivlen o he ringle inequliy! To pu his noher wy, le l 2 (x) = x 2 + x 2 2, x = (x, x 2 ) R 2, hen l 2 (x + y) l 2 (x) + l 2 (y), x, y R 2 In oher words, l 2 is subddiive Bu i s lso homogeneous of degree one Hence, i is convex on R 2, ie, x, y R 2, nd 0 λ, hen l 2 (λx + ( λ)y) λl 2 (x) + ( λ)l 2 (y) To ge n ide of wh his mens, geomericlly, ry nd picure he region of poins (x, y, z) in 3spce consrined by he condiions 0 z x 2 + y 2, x, y R [Hin: your nex mel exmine he sugr bowl, sy] A soluion of Problem follows once we cn show h l 3 (x) = x 2 + x x 2 3, x = (x, x 2, x 3 ) R 3, hs he sme propery Exercise 5 Esblish h l 3 is subddiive nd convex on R 3, nd solve Problem 6 Soluion of Problem 3 By now i will hve occurred o you h wh is wned o solve Problem 3 is n nlogue of Corollry for he hrmonic men of hree posiive numbers (This hs properies similr o he geomeric men, bu doesn ge s much enion, nd doesn feure in oo mny IMO problems) Nmely, we wn o show h he funcion h 3 defined by 3 h 3 (x) = x + x 2 +, x = (x, x 2, x 3 ) R 3, x, x 2, x 3 > 0 x 3 is superddiive, nd so concve, on R 3 Theorem 7 If, b, c > 0, hen + + = min{x + by + cz : 0 < x, y, z, x + y + z = } b c 6
7 Proof This is consequence of Theorem 4 For, if x, y, z > 0 nd x+ y+ z =, hen Hence, = ( x + by b + cz c ) 2 (x + by + cz)( + b + c ) + + x + by + cz, b c whenever 0 < x, y, z, x + y + z =, nd here is equliy here when x = 2 ( + +, y = b c )2 b 2 ( + +, z = b c )2 c 2 ( + + b c )2 Corollry 3 If, b, c, d, e, f > 0, hen h 3 (, b, c) + h 3 (d, e, f) h 3 ( + d, b + e, c + f) There is equliy iff d = b e = c f Proof For, if x, y, z > 0 nd x + y + z =, hen h 3 (, b, c)+h 3 (d, e, f) 3(x+by+cz)+3(dx+ey+fz) = 3[(+d)x+(b+e)y+(c+f)z], nd so h 3 (, b, c) + h 3 (d, e, f) doesn exceed 3 min{(+d)x+(b+e)y+(c+f)z : 0 < x, y, z, x+ y+ z = } = h 3 (+d, b+e, c+f) Moreover, (?) here is equliy here iff nd x = ( + + ) = d( + + ) = ( + d)( + + ), b c d e f +d b+e c+f y = b( + + ) = e( + + ) = (b + e)( + + ), b c d e f +d b+e c+f z = c( + + ) = f( + + ) = (c + f)( + + ) b c d e f +d b+e c+f In oher words, (?) iff Exercise 6 Now solve Problem 3 d = b e = c f 7
8 7 Exercises Suppose, b > 0 Prove h = b + g 2 (, b) 2 g 2 (, + b 2 ), 2 Se nd prove n nlogue of Theorem 7 nd is corollry for h 2, where 3 Suppose, b > 0 Prove h 2 h 2 (x) = x +, x = (x, x 2 ) R 2, x i, x 2 > 0 x 2 = b + h 2 (, b) 2 h 2 (, + b 2 ), 4 Suppose, b, c > 0 Prove h g 2 (, b) + g 2 (b, c) + g 2 (c, ) + b + c, = b = c 5 Suppose, b, c > 0 Prove h h 2 (, b) + h 2 (b, c) + h 2 (c, ) + b + c, = b = c 6 Suppose, b, c > 0 Prove h 3 3 bc = min{x + by + cz : 0 < x, y, z, xyz = } Deduce Corollry 7 More generlly, if x i 0, i =, 2,, n, x = (x, x 2,, x n ), nd g n (x) = n x x 2 x n Prove h n ng n (x) = min{ x i y i : y y 2 y n =, y i > 0, i =, 2, n} i= Now redo Exercise 2 in Secion 2 8
9 8 Suppose, b, c > 0 Prove h = b = c + g 2 (, b) + g 3 (, b, c) 3g 3 (, + b 2, + b + c ), 3 9 Suppose, b, c > 0 Prove h + h 2 (, b) + h 3 (, b, c) 3h 3 (, + b 2, + b + c ), 3 = b = c 0 Suppose, b, c, d > 0 Prove h b + b + + cd c + d + Suppose, b, c > 0 Prove h < ( + c)(b + d) + b + c + d + g 2 (, b) + g 2 (b, c) + g 2 (c, ) 3g 3 ( + b 2, b + c 2, c + 2 ) + b + c, = b = c 2 Suppose, b, c > 0 Prove h h 2 (, b) + h 2 (b, c) + h 2 (c, ) 3h 3 ( + b 2, b + c 2, c + 2 ) + b + c, = b = c 9
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