SolFeb2205 argz

Claim: If $a, b, c$ are positive reals, then $a^a b^b c^c \ge (abc)^{(a+b+c)/3}$ .

Proof: (Due to argz.) Assume, wlog, that $a \le b \le c$ . Then, since $a \le b$

$\left(\frac{a}{b}\right)^a \ge \left(\frac{a}{b}\right)^b ,$

so $a^ab^b \ge a^b b^a$ . Hence by the same reasoning,

$(a^a b^b c^c)^2 = (a^ab^b)(a^ac^c)(b^bc^c) \ge (a^b b^a)(a^c c^a)(b^c)(c^b)= a^{b+c}b^{a+c}c^{a+b},$

and multiplying both sides by $a a b b c c$ and taking cube roots gives us the inequality.