因為 \(a,b,c\in\mathbb{R^{+}},\)
\(\displaystyle \frac{a^8}{8}+\frac{a^8}{8}+\frac{b^8}{8}+\frac{b^8}{8}+\frac{b^8}{8}+\frac{c^8}{8}+\frac{c^8}{8}+\frac{c^8}{8}\geq 8 \sqrt[8]{\left(\frac{a^8}{8}\right)^2\left(\frac{b^8}{8}\right)^3\left(\frac{c^8}{8}\right)^3}=a^2b^3c^3,\)
\(\displaystyle \frac{a^8}{8}+\frac{a^8}{8}+\frac{a^8}{8}+\frac{b^8}{8}+\frac{b^8}{8}+\frac{c^8}{8}+\frac{c^8}{8}+\frac{c^8}{8}\geq 8 \sqrt[8]{\left(\frac{a^8}{8}\right)^3\left(\frac{b^8}{8}\right)^2\left(\frac{c^8}{8}\right)^3}=a^3b^2c^3,\)
\(\displaystyle \frac{a^8}{8}+\frac{a^8}{8}+\frac{a^8}{8}+\frac{b^8}{8}+\frac{b^8}{8}+\frac{b^8}{8}+\frac{c^8}{8}+\frac{c^8}{8}\geq 8 \sqrt[8]{\left(\frac{a^8}{8}\right)^3\left(\frac{b^8}{8}\right)^3\left(\frac{c^8}{8}\right)^2}=a^3b^3c^2,\)
以上三式相加，可得 \(\displaystyle a^8+b^8+c^8 \geq a^2 b^3 c^3 + a^3 b^2 c^3 + a^3 b^3 c^2\)
\(\displaystyle \Leftrightarrow \frac{a^8+b^8+c^8}{a^3b^3c^3}\geq\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
