<s>[INST] 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?
 [/INST] Let $a＝x^2,b＝3y^2$ and $t＝\frac{x}{y}$,only need to find the maxium of this
$$\frac{7(t^2+3t+6)^2}{2t^4+21t^2+2}$$
and ...

</s>