
A2. Let $a$ and $b$ be positive real numbers such that $3 a^{2}+2 b^{2}=3 a+2 b$. Find the minimum value of

$$
A=\sqrt{\frac{a}{b(3 a+2)}}+\sqrt{\frac{b}{a(2 b+3)}}
$$


Solution. By the Cauchy-Schwarz inequality we have that

$$
5\left(3 a^{2}+2 b^{2}\right)=5\left(a^{2}+a^{2}+a^{2}+b^{2}+b^{2}\right) \geq(3 a+2 b)^{2}
$$

(or use that the last inequality is equivalent to $(a-b)^{2} \geq 0$ ).

So, with the help of the given condition we get that $3 a+2 b \leq 5$. Now, by the AM-GM inequality we have that

$$
A \geq 2 \sqrt{\sqrt{\frac{a}{b(3 a+2)}} \cdot \sqrt{\frac{b}{a(2 b+3)}}}=\frac{2}{\sqrt[4]{(3 a+2)(2 b+3)}}
$$

Finally, using again the AM-GM inequality, we get that

$$
(3 a+2)(2 b+3) \leq\left(\frac{3 a+2 b+5}{2}\right)^{2} \leq 25
$$

so $A \geq 2 / \sqrt{5}$ and the equality holds if and only if $a=b=1$.
