Calculate the minimum value of $\frac{a}{\sqrt{a + 2b}} + \frac{b}{\sqrt{b + 2a}}$ where $a, b > 0$ and $\sqrt{a + 2b} = 2 + \sqrt{\frac{b}{3}}$.

Question

Given that $a$ and $b$ are positives such that $\sqrt{a + 2b} = 2 + \sqrt{\dfrac{b}{3}}$, calculate the minimum value of $$\large \frac{a}{\sqrt{a + 2b}} + \frac{b}{\sqrt{b + 2a}}$$

I have provided two solutions, one comes out of stupidity and the other is a result of guesstimates. There should and must be better solutions.

Answer 1

We have that $$\frac{a}{\sqrt{a + 2b}} + \frac{b}{\sqrt{b + 2a}} + \sqrt{\frac{b}{3}} \ge \frac{(a + b + b)^2}{a\sqrt{a + 2b} + b\sqrt{b + 2a} + b\sqrt{3b}}$$

$$\ge \frac{(a + 2b)^2}{a\sqrt{a + 2b} + b\sqrt{(1 + 1)(b + 2a + 3b)}} = \frac{(a + 2b)^2}{a\sqrt{a + 2b} + 2b\sqrt{a + 2b}} = \frac{(a + 2b)^2}{(a + 2b)\sqrt{a + 2b}} = \sqrt{a + 2b}$$

$$\implies \frac{a}{\sqrt{a + 2b}} + \frac{b}{\sqrt{b + 2a}} \ge \sqrt{a + 2b} - \sqrt{\frac{b}{3}} = 2$$

The equality sign occurs when $a = b = 3$.

Answer 2

Let $\left\{ \begin{align} x &= \sqrt{a + 2b}\\ y &= \sqrt{\dfrac{b}{3}} \end{align} \right.$$\iff \left\{ \begin{align} x - y &= 2\\ a &= x^2 - 6y^2\\ b &= 3y^2 \end{align} \right.$. That means $$\frac{a}{\sqrt{a + 2b}} + \frac{b}{\sqrt{b + 2a}} = \frac{x^2 - 6y^2}{x} + \frac{3y^2}{\sqrt{2x^2 - 9y^2}} = \frac{(x^2 - 6y^2)\sqrt{2x^2 - 9y^2} + 3xy^2}{x\sqrt{2x^2 - 9y^2}}$$

$$ = \frac{(x^2 - 6y^2)\sqrt{2x^2 - 9y^2} - x(x - y)\sqrt{2x^2 - 9y^2} + 3xy^2}{x\sqrt{2x^2 - 9y^2}} + 2$$

$$ = \frac{y \cdot \left[(x - 6y)\sqrt{2x^2 - 9y^2} + 3xy\right]}{x\sqrt{2x^2 - 9y^2}} + 2 = y \cdot \left[1 - \frac{6y}{x} + \frac{3y}{\sqrt{2x^2 - 9y^2}}\right] + 2$$

Furthermore, $(a - b)^2 \ge 0, \forall a, b \in \mathbb R \implies (a + 2b)^2 \ge 3b(b + 2a), \forall a, b \in \mathbb R$

$\iff a + 2b \ge \sqrt{3b(b + 2a)} \iff x \ge 3y\sqrt{2x^2 - 9y^2} \iff \dfrac{3y}{\sqrt{2x^2 - 9y^2}} \ge \dfrac{9y^2}{x^2}$

$$y \cdot \left[1 - \frac{6y}{x} + \frac{3y}{\sqrt{2x^2 - 9y^2}}\right] + 2 \ge y \cdot \left(1 - \frac{3y}{x}\right)^2 + 2 \ge 2$$

This equality sign occurs when $a = b = 3$.

Answer 3

We can kill it by the direct substitution!

For $a=b=3$ we obtain a value $2$.

We'll prove that it's a minimal value.

Indeed, let $\sqrt{\frac{b}{3}}=x$.

Thus, the condition gives $a=4-5x^2+4x$ and we need to prove that $$\frac{4-5x^2+4x}{2+x}+\frac{3x^2}{\sqrt{8-7x^2+8x}}\geq2$$ or $$\frac{3x}{\sqrt{8-7x^2+8x}}\geq\frac{5x-2}{2+x}.$$ Now, if $5x-2\leq0$ and $8-7x^2+8x>0$ so our inequality is obviously true.

But for $5x-2>0$ and $8-7x^2+8x>0$ it's enough to prove that $$9x^2(2+x)^2\geq(5x-2)^2(8-7x^2+8x)$$ or $$(x-1)^2(23x^2+8x-4)\geq0,$$ which is obvious.