Let's Play AM GM

3x + 4y=5 So, 3x/2 + 3x/2 + 4y/3 + 4y/3 + 4y/3 = 5. Apply AM-GM inequality on the Left Hand Side,

LHS >= (3x/2 . 3x/2 . 4y/3 . 4y/3 .4y/3)=(16/3 .x^2.y^3)^{1/5}

So, 1>= 16/3 .x^2 .y^3 So, x^2.y^3 <= 3/16

So, a=3, b=16 So a+b =19.

$3x+4y=5$ so $x=\frac{5-4y}{3}$ . Substitute in next line to get $(\frac{5-4y}{3})^2y^3\leq \frac{a}{b}$ .

Maximize by taking first derivative and setting equal to zero.

$3y^2(\frac{5-4y}{3})^2+ 2y^3(\frac{5-4y}{3})*\frac{-4}{3} =0$

$y^2(\frac{5-4y}{3})( 3*\frac{5-4y}{3} + \frac{-8y}{3})=0$

$y^2(\frac{5-4y}{3})( \frac{15-20y}{3})=0$

$y=0, \frac{5}{4},$ or $\frac{3}{4}$ .

If $y=0,$ and $x= \frac{5}{3}$ then $x^2y^3=0$

If $y=\frac{5}{4},$ and $x=0$ then $x^2y^3=0$

If $y=\frac{3}{4},$ and $x=\frac{2}{3}$ then $x^2y^3=(\frac{2}{3})^2(\frac{3}{4})^3 = \frac{3}{16}$ .

$a=3,b=16, a+b=19$

As $x,y \in \mathbb R^+$ , by weighted $AM-GM$ inequality, we have:

$\frac{2 \cdot (\frac{3x}{2}) + 3 \cdot (\frac{4y}{3})}{2+3} \geq (\frac{9x^2}{4} \cdot \frac{64y^3}{27})^\frac{1}{5}$

$\Rightarrow (1)^5 \geq \frac{16}{3} x^2y^3$

$\Rightarrow \frac{3}{16} \geq x^2y^3$