To show that x^{2}+y^{2}+z^{2} <= x^{3}+y^{3}+z^{3} is to show that f = x^{3}+y^{3}+z^{3} - (x^{2}+y^{2}+z^{2}) >= 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 = x^{3}+y^{3}+z^{3} - (x^{2}+y^{2}+z^{2})
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(t^{2}+r^{2}+s^{2}) + t^{3}-r^{3}-s^{3})
- (3 + 2(t-r-s) + t^{2}+r^{2}+s^{2})
f = t-r-s + 2(t^{2}+r^{2}+s^{2}) + t^{3}-r^{3}-s^{3}
Let g = 2(t^{2}+r^{2}+s^{2})
f = r/x + s/y + rs/xy - r - s + g + (r/x + s/y + rs/xy)^{3}-r^{3}-s^{3}
(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 - r^{3} - s^{3}
f = (r/x - r) + (s/y - s) + rs/xy + g + ((r/x)^{3} - r^{3}) + ((s/y)^{3} - s^{3}) + h
f = r(1/x - 1) + s(1/y - 1) + rs/xy + g + r^{3}((1/x)^{3} - 1) + s^{3}((1/y)^{3} - 1) + h
f = r(1 + r/x - 1) + s(1 + s/y - 1) + rs/xy + g + r^{3}((1 + r/x)^{3} - 1) + s^{3}((1 + s/y)^{3} - 1) + h
f = r(r/x) + s(s/y) + rs/xy + g + r^{3}((1 + r/x)^{3} - 1) + s^{3}((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.