Prove that $\frac{a}{2a+\beta b}+\frac{b}{\alpha b+\beta a}\ge \frac{2}{\alpha +\beta }$

Question

Let $a;b;\alpha;\beta>0$ and $\beta>\alpha $. Prove that $$\frac{a}{2a+\beta b}+\frac{b}{\alpha b+\beta a}\ge \frac{2}{\alpha +\beta }$$

---

$$LHS-RHS=\frac{a^2\alpha \beta+a^2\beta^2-4a^2\beta+ab\alpha^2+ab\alpha\beta-2ab\alpha-2ab\beta^2+2ab\beta-b^2\alpha\beta+b^2\beta^2}{(\alpha +\beta)(2a+\beta b)(\alpha b+\beta a) }\ge0$$

Then i can not solve it, help me !

Answer 1

It's wrong.

Try $\beta=1$, $\alpha=\frac{1}{2}$ and $a=b=1.$