<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] The mistake is when you apply AM-GM in the last step. Applying it you get $\ge$ inequality, which corresponds to the minimal value of l.h.s, not maximal. 

Also $m \le \frac{7}{2\sqrt{3ab}} + \sqrt{3ab}$, not equal. 
UPD: You did the following 
$m = \frac{7}{a + 3b} + \sqrt{3ab} \le \frac{7}{2\sqrt{3ab}} + \sqrt{3ab} \ge 2\sqrt{\frac 72}$ and that's it. You get $2\sqrt{\frac 72}$ as maximal value. 
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