Question is 

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> For positive numbers $a,b$ such that $2a^2 +7ab+3b^2 = 7$, what is the maximum value of $a+{\sqrt{3ab}}+2b$?
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I use AM-GM Inequality to do this

$$(2a+b)(a+3b)=7\text{ and }2a+b=\frac{7}{a+3b}.$$

So, maximum value $m$ is $\frac{7}{a+3b}+{\sqrt{3ab}}$ so $m=\frac{7}{2{\sqrt{3ab}}}+{\sqrt{3ab}}$ is $2{\sqrt{\frac{7}{2}}}$ so $2m^2=28$.

But it didn't satisfy equal condition so how do I get it?