Bump....

I think it should be $(2b+3)$ instead of $(3b+2)$ By Cauchy: $(3a^2+2b^2)(3+2) \geq (3a+2b)^2$ so $5 \geq (3a+2b)$. Now we apply $AM-GM$ on $A$: $A\geq 2 \sqrt[4]{\frac{ab}{ab(3a+2)(2b+3)}}$ And by $AM-GM$ on $\sqrt[4]{(3a+2)(2b+3)}\leq \sqrt[4]{\frac{1}{4}(3a+2b+5)^2} \leq \sqrt[4]{25}$ so $A\geq \frac {2}{\sqrt{5}}$ (equality when $a=b=1$)

Steve12345 wrote: I think it should be $(2b+3)$ instead of $(3b+2)$ ... you are right, I have a typo, thanks for noticing, I shall correct the last fraction, this belowe was the original post with the typo parmenides51 wrote: Let $a$ and $b$ be positive real numbers such that $3a^2 + 2b^2 = 3a + 2b$. Find the minimum value of $A =\sqrt{\frac{a}{b(3a+2)}} + \sqrt{\frac{b}{a(3b+2)}} $

This problem was proposed by me.

Let $a$ and $b$ be positive real numbers such that $3a^2 + 2b^2 = 3a + 2b$. Then$$\sqrt{\frac{a}{b(3a+2)}} + \sqrt{\frac{b}{a(2b+3)}} \geq \frac {2}{\sqrt{5}},$$

By AM-GM minimal value $A\ge 2{\sqrt [4]{\frac{1}{(3a+2)(3b+2)}}}\ge\frac {2}{\sqrt 5}\Leftrightarrow (3a+2)(3b+2)\le 25$ Now, $3\left(a-\frac {1}{2}\right)^2+2\left (b-\frac {1}{2}\right)^2=\frac {5}{4}\Leftrightarrow 3 (2a-1)^2+2 (2b-1)^2=5$ for $a=b=1$, and $A=\frac {2}{\sqrt 5} $ Consider $(2a-1)=c $, $(2b-1)=d $ Therefore, $3c^2+2d^2=5$ We need to prove that $100\ge(3x+7)(2y+7)$ By C-S $5(3c^2+2d^2)\ge (3c+2d)^2\Leftrightarrow 5\ge (3c+2d) $ Case 1. $1\ge d$ $(3c+7)\cdot(2d+7)\le\left (\frac {3c+2d+14}{2}\right)^2=100$ Case 2. $d>1$ Let $d=1+x $, $x>0$ Note, $3c^2+2x^2+4x=3$ $100\ge (3c+7)\cdot (3x+10)\Leftrightarrow 1-\frac {10x}{10+3x}\ge c$ A contradiction $c>1-\frac {10x}{10+3x}$ $3>3\left (1-\frac {10x}{10+3x}\right)^2+2x^2+4x\Leftrightarrow 0> 90x^3+78x^2+240x+50$. It is simply not possible. Only the equality will hold when $a=b=1$ and $c=d=1$

sqing wrote:

Let $a$ and $b$ be positive real numbers such that $3a^2 + 2b^2 = 3a + 2b$. Then$$\sqrt{\frac{a}{b(3a+2)}} + \sqrt{\frac{b}{a(2b+3)}} \geq \frac {2}{\sqrt{5}},$$ No offence Physicsknight, but Easier(or, less bashier rather) solution here From given condition, by QM-AM, we get $3a+2b\leq 5\implies a\leq \frac{5-2b}{3}$ Now, by AM-GM, $\sqrt{\frac{a}{b(3a+2)}} + \sqrt{\frac{b}{a(2b+3)}} \geq \frac{2}{\sqrt[4]{(6ab+9a+4b+6)}}$ So, we are left to prove $6ab+9a+4b+6\leq 25\Leftrightarrow a(6b+9)+4b\leq 19\Leftrightarrow a\leq \frac{19-4b}{6b+9}$. But, $a\leq \frac{5-2b}{3}$ So, we are left to prove $\frac{5-2b}{3}\leq \frac{19-4b}{6b+9}\Leftrightarrow (b-1)^2\geq 0$ Equality holds when $a=b=1$ Nice problem, silouan

By Holder, we have \[\left(\sqrt{\frac{a}{b(3a+2)}} + \sqrt{\frac{b}{a(2b+3)}}\right)^2 \left(a^2b(3a+2)+ab^2(2b+3)\right) \geq (a+b)^3\]and so \[\left(\sqrt{\frac{a}{b(3a+2)}} + \sqrt{\frac{b}{a(2b+3)}}\right)^2 \geq \frac{(a+b)^3}{ab(3a^2+2a+2b^2+3b)}=\frac{(a+b)^3}{5ab(a+b)}\geq \frac{4ab}{5ab}=\frac{4}{5}\]thus, $A\geq \frac{2}{\sqrt{5}}$ equality holds when $a=b=1$.

The answer is $\frac{2}{\sqrt 5}$. Equality is achieved when $a=b=1$. We now show this is a bound. By Holder, we have that $(\sqrt{\frac{a}{b(3a+2)}}+\sqrt{\frac{b}{a(2b+3)}}^2)(a^2b(3a+2)+b^2a(2b+3)\geq (a+b)^3$. Notice that multiplying both sides of the given condition by $ab$ gives that $3a^3b+2ab^3=3a^2b+2ab^2,$ and substituting this, we get that $(\sqrt{\frac a {b(3a+2)}}+\sqrt{\frac{b}{a(2b+3)}})^2\geq \frac{(a+b)^2}{5ab}$. Using AM-GM finishes.