How many ways can you solve this?

I call the expression is A

Without losing generality we assume that $a\geq b\geq c$

Therefore $a^2\geq b^2\geq c^2$ and $\frac{1}{b+2c}\geq\frac{1}{c+2a}\geq\frac{1}{a+2b}$

Applying Chebyshev's Inequality we get $3A\geq 16(a^2+b^2+c^2)\bigg(\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\bigg)$ $\Leftrightarrow A\geq 16\bigg(\frac{3}{a+b+c}\bigg)$ Since $a+b+c\leq\sqrt{3(a^2+b^2+c^2)}=3$ , the minimum value of A's 16

*Note: You can also apply Holder's Inequality to get the answer

It's to easy , a , b , c are real +be no. Then let a=b=c=1, ( always let as it satisfy given condition) . Now put a,b,c in eqn to get answer.