For positive number $a,b$, when $a,b$ satisfies $2a^2 +7ab+3b^2=7$, what is maximum value of $a+{\sqrt{3ab}}+2b$ Question is 

For positive numbers $a,b$ such that $2a^2 +7ab+3b^2 = 7$, what is the maximum value of $a+{\sqrt{3ab}}+2b$?

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?

Your answer is right!
Indeed, we'll prove that $\sqrt{14}$ is a maximal value.
Let $a=x^2$ and $b=3y^2$, where $x$ and $y$ are positives.
Thus, $2a^2+7ab+3b^2=2x^4+21x^2y^2+27y^4$ and we need to prove that
$$x^2+3xy+6y^2\leq\sqrt{14}$$ or
$$(x^2+3xy+6y^2)^2\leq2(2x^4+21x^2y^2+27y^4)$$ or
$$(x-y)^2(x^2+6y^2)\geq0,$$ which is obvious.
The equality occurs for $b=3a$, which says that we got a maximal value.
