I just came across the following problem:
Let $a, b, c$ be positive real numbers such that $abc=1$. Prove that:
$(a^5+a^4+a^3+a^2+a+1)(b^5+b^4+b^3+b^2+b+1)(a^5+a^4+a^3+a^2+a+1)\ge 8(a^2+a+1)(b^2+b+1)(c^2+c+1)$
I proved it in the following fashion:
$(a^5+a^4+a^3+a^2+a+1)(b^5+b^4+b^3+b^2+b+1)(a^5+a^4+a^3+a^2+a+1)=(a^3+1)(b^3+1)(c^3+1)(a^2+a+1)(b^2+b+1)(c^2+c+1)$
So we just have to prove that $(a^3+1)(b^3+1)(c^3+1)\ge 8$, which is true from the following:
$(a^3+1)(b^3+1)(c^3+1)=(a^3b^3+a^3+b^3+1)(c^3+1)=a^3b^3c^3+1+a^3b^3+a^3c^3+b^3c^3+a^3+b^3+c^3$
$=2+a^3b^3+a^3c^3+b^3c^3+a^3+b^3+c^3$
$\ge2+3\sqrt[3]{a^6b^6c^6}+3\sqrt[3]{a^3b^3c^3}$
$=8$
So it holds true, so the inequality is proved.
I am wondering what other solutions exist to this problem, could you please post up alternative approaches?
Best Answer
From your idea, we need to prove $$(a^3+1)(b^3+1)(c^3+1) \geqslant 8.$$ Proof 1. Using the AM-GM inequality we have $$(a^3+1)(b^3+1)(c^3+1) \geqslant 2\sqrt{a^3} \cdot 2\sqrt{b^3} \cdot 2\sqrt{c^3} = 8\sqrt{a^3b^3c^3} = 8.$$ Proof 2. According to the Holder inequality, we have $$(a^3+1)(b^3+1)(c^3+1) \geqslant \big(\sqrt[3]{a^3b^3c^3}+1\big)^3 = (abc+1)^3 = 8.$$