Solution May 30, 2007

Problem

Let a,b,c,d be positive reals such that a + b + c + d = 1.

Prove that: 6(a^{3}+b^{3}+c^{3}+d^{3})\geq a^{2}+b^{2}+c^{2}+d^{2}+\frac{1}{8}.

Solution

Let S_1=\frac{a+b+c+d}4, S_2=\frac{a^2+b^2+c^2+d^2}4 and S_3=\frac{a^3+b^3+c^3+d^3}4. By the power mean inequality (http://planetmath.org/encyclopedia/GeneralMeansInequality.html), we have

S_1 \leq \sqrt{S_2} \leq \sqrt[3]{S_3}.

This implies

24 S_3 \geq 16 S_2 S_1 + 8 S_1^3

Using S_1 = \frac 14, this is equivalent to

6(a^{3}+b^{3}+c^{3}+d^{3})\geq a^{2}+b^{2}+c^{2}+d^{2}+\frac{1}{8}

q.e.d.