parmenides51 wrote: Positive real numbers $a, b$ are such that $a^3 + b^3 = 2$. Show that that $\frac{1}{a}+\frac{1}{b}\ge 2(a^2 - a + 1)(b^2 - b + 1)$. Solution of Zhangyanzong: By AM-GM, $$\frac{(a+b)(a+1)(b+1)}{ab}\ge 8=\frac{(a^3 + b^3 + 2)^2}{2}\geq2(a^3 + 1)(b^3 + 1)$$$$\frac{1}{a}+\frac{1}{b}\ge\frac{8}{(a+1)(b+1)}\ge 2(a^2 - a + 1)(b^2 - b + 1)$$

Positive real numbers $a, b, c$ are such that $a^3 + b^3+ c^3 = 3$. Show that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{81}{(a+2)(b+2)(c+2)}\geq \frac{1}{9}(a^2 - 2a + 4)(b^2 - 2b + 4)(c^2 - 2c + 4)$$Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be positive real numbers such that $a_1^3+a_2^3+\cdots+a_n^3=n.$ Prove that$$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\geq \frac{n^{n+1}}{(a_1+n-1)(a_2+n-1)\cdots (a_n+n-1)}$$$$\geq \frac{n}{(n^2-3n+3)^n}(a^2_1 -(n-1) a_1 + (n-1)^2)(a^2_2 - (n-1)a_2 + (n-1)^2)\cdots(a^2_n - (n-1)a_n + (n-1)^2).$$(Zhangyanzong)