Proving :$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$

Question

Let $a,b,c>0$ be real numbers such that $a+b+c=3$,how to prove that? :

$$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$$

Answer 1

By AM>GM $$ \frac{a+b+c}{3}=1 \Rightarrow abc\le 1 \\ \Rightarrow \frac{1}{2ab^2+1} = \frac{1}{2abc\frac{b}{c}+1}\ge\frac{1}{2\frac{b}{c}+1}=\frac{c}{2b+c} \\ S = \frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge\frac{c}{2b+c}+\frac{a}{2c+a}+\frac{b}{2a+b} $$ $$ \begin{align} 3S-3 & \ge \frac{3c-2b-c}{2b+c}+\frac{3a-2c-a}{2c+a}+\frac{3b-2a-b}{2a+b}\\ & = 2\left(\frac{c-b}{2b+c}+\frac{a-c}{2c+a}+\frac{b-a}{2a+b}\right) \\ & = \frac{2}{D}\left(3ab^2+3bc^2+3ca^2-9abc\right)\\ & = \frac{6abc}{D}\left(\frac{b}{c}+\frac{c}{a}+\frac{a}{b}-3\right) \\ & \ge 0 \end{align} $$ where for clarity we simply write $D$ for the positive denominator, and the last inequality is again by AM>GM $$ 1=\left(\frac{b}{c}\cdot\frac{c}{a}\cdot\frac{a}{b}\right)^{1/3}\le\frac{1}{3}\left(\frac{b}{c}+\frac{c}{a}+\frac{a}{b}\right) $$ Finally $3S-3\ge 0 \Rightarrow S\ge 1$.