maximum value of $a^2b$ condition is given

Question

the maximum value of $a^2b$ subjected to the condition $a+b+\sqrt{2a^2+2ab+3b^2}=10$ given $a,b\geq 0$

solution i try

$\displaystyle \frac{a}{2}+\frac{a}{2}+b\geq \left(\frac{a^2}{4}b\right)^{\frac{1}{3}}$

for $2a^2+2ab+3b^2=2(a+b)^2-ab+b^2$

$a+b+\sqrt{2a^2+2ab+3b^2}=(a+b)+\sqrt{(a+b)^2-ab+b^2}\geq \left(\frac{a^2}{4}b\right)^{\frac{1}{3}}$

How do i solve it from here

Answer 1

The condition gives $$2a^2+2ab+3b^2=(10-a-b)^2$$ or $$a^2+2b^2+20(a+b)=100$$ or $$(a+10)^2+2(b+5)^2=250$$ or $$2\left(\frac{a}{2}+5\right)^2+(b+5)^2=125.$$ Now, by AM-GM and Holder we obtain: $$125=2\left(\frac{a}{2}+5\right)^2+(b+5)^2\geq3\sqrt[3]{\left(\frac{a}{2}+5\right)^4(b+5)^2}=$$ $$=3\left(\sqrt[3]{\left(\frac{a}{2}+5\right)^2(b+5)}\right)^2\geq3\left(\sqrt[3]{\left(\sqrt[3]{\frac{a^2b}{4}}+5\right)^3}\right)^2=3\left(\sqrt[3]{\frac{a^2b}{4}}+5\right)^2.$$ Id est, $$a^2b\leq500\left(\sqrt{\frac{5}{3}}-1\right)^3.$$ The equality occurs for $b=\frac{a}{2}$, which says that we got a maximal value.