[Solution] Sum of squares <= Sum of cubes.

Problem was to show that

$$a^2 + b^2 + c^2 \le a^3 + b^3 + c^3$$

if $a,b,c, \gt 0$ and $abc = 1$.


Let us first show that

$$a^{a^2} b^{b^2} c^{c^2} \ge 1$$

If $a \ge b \ge 1 \ge c$ then we rewrite using $c = \frac{1}{ab}$ as


$$a^{a^2 - c^2} b^{b^2 - c^2}$$ which is easily seen to be at least $1$ (product of numbers $\ge 1$).

If $a \ge 1 \ge b \ge c$ then we rewrite using $a = \frac{1}{bc}$ as


$$b^{b^2 - a^2} c^{c^2 - a^2} \ge 1$$

(product of numbers less than $1$ raised to negative powers)


Now for the main problem, we can use the weighted arithmetic mean geometric mean inequality

and get


$$\frac{a.a^2 + b.b^2 + c.c^2}{a^2 + b^2 + c^2} \ge \sqrt[a^2+b^2+c^2]{a^{a^2}b^{b^2}c^{c^2}}$$

But RHS $\ge 1$ and so we are done.