beero1000
Veteran Member
Assume without loss of generality that x <= y <= z. z = 1/xy. If all three were less than 1 their product would be less than 1; likewise, if all three were greater than 1 their product would be greater than 1. Therefore x <= 1 and z >= 1. There are two cases to consider, y <= 1 and y > 1. Here's a proof for y <= 1.1. If x,y,z are positive real numbers with xyz = 1, show that x2+y2+z2 ≤ x3+y3+z3
To show that x2+y2+z2 <= x3+y3+z3 is to show that f = x3+y3+z3 - (x2+y2+z2) >= 0.
Let r = 1 - x, s = 1 - y, t = z - 1. r, s, t >= 0.
x(1 + r/x) = x + r = x + 1 - x = 1. Therefore 1 + r/x = 1/x. Similarly, 1 + s/y = 1/y.
z = 1/xy = (1/x)(1/y) = (1 + r/x)(1 + s/y) = 1 + r/x + s/y + rs/xy
t = r/x + s/y + rs/xy
f = x3+y3+z3 - (x2+y2+z2)
f = (1-r)3+(1-s)3+(1+t)3 - ((1-r)2+(1-s)2+(1+t)2)
f = (3 + 3(t-r-s) + 3(t2+r2+s2) + t3-r3-s3)
- (3 + 2(t-r-s) + t2+r2+s2)
f = t-r-s + 2(t2+r2+s2) + t3-r3-s3
Let g = 2(t2+r2+s2)
f = r/x + s/y + rs/xy - r - s + g + (r/x + s/y + rs/xy)3-r3-s3
(r/x + s/y + rs/xy)3 = (r/x)3 + (s/y)3 + [a series of 25 more products of sets of three terms].
Let h = that series of 25 products I'm not going to painfully write out.
f = r/x + s/y + rs/xy - r - s + g + (r/x)3 + (s/y)3 + h - r3 - s3
f = (r/x - r) + (s/y - s) + rs/xy + g + ((r/x)3 - r3) + ((s/y)3 - s3) + h
f = r(1/x - 1) + s(1/y - 1) + rs/xy + g + r3((1/x)3 - 1) + s3((1/y)3 - 1) + h
f = r(1 + r/x - 1) + s(1 + s/y - 1) + rs/xy + g + r3((1 + r/x)3 - 1) + s3((1 + s/y)3 - 1) + h
f = r(r/x) + s(s/y) + rs/xy + g + r3((1 + r/x)3 - 1) + s3((1 + s/y)3 - 1) + h
By inspection, every individual term in the above sum of terms is non-negative. Therefore f >= 0. Q.E.D.
The proof for y > 1 is analogous and is left as an exercise for the masochist.
Ouch!