Prove $\frac{a}{ab+2c}+\frac{b}{bc+2a}+\frac{c}{ca+2b} \ge \frac 98$ $a,b,c \in \mathbb{R^+} \text{such that }a+b+c=2$. Prove inequality $$\frac a{ab+2c}+\frac b{bc+2a}+\frac c{ca+2b} \ge \frac 98$$

I tried


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*$$LHS = \sum \frac{1}{b+2\cdot c/a} \ge \frac 9 {2+2(\sum c/a)} \longrightarrow failed$$

*$$\frac a {ab+2c} \ge \frac 9{16}a \longrightarrow failed$$


\begin{align}
\sum_{cyc}{\frac{a}{ab+2c}}=\sum_{cyc}{\frac{a}{ab+(a+b+c)c}}=\sum_{cyc}{\frac{a}{(a+c)(b+c)}}& =\frac{\sum_{cyc}{a(a+b)}}{(a+b)(a+c)(b+c)} \\
& =\frac{a^2+b^2+c^2+ab+ac+bc}{(a+b)(a+c)(b+c)} \\
& \geq \frac{\frac{2}{3}(a+b+c)^2}{(a+b)(a+c)(b+c)} \\
& \geq \frac{\frac{2}{3}(a+b+c)^2}{\left(\frac{(a+b)+(a+c)+(b+c)}{3}\right)^3} \\
& =\frac{9}{8}
\end{align}
