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

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?

Answer 1

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.

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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.

Answer 2

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.

Answer 3

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 ...