What's the lowest value of $\sqrt{1+\frac4x} - \sqrt{1-x}$ for $0 < x ≤ 1$

Question

What's the lowest value of $\sqrt{1+\frac4x} - \sqrt{1-x}$ for $0 < x ≤ 1$

Encountered this on 2018 Tubitak National Mathematics Olympiads and tried to solve it using Arithmetic and Geometric Mean Inequality but couldn't manage to get an answer, would be appreciated if anyone could solve using AGM (other solutions would be appreciated too if it's not too advanced).

Answer 1

Let $\varphi(x) = \sqrt{1+\frac4x} - \sqrt{1-x}$.

For $x \in (0,1]$, $\varphi(x) \ge \sqrt{1+\frac41} - 1 = \sqrt{5} - 1 > 0$. This means minimizing $\varphi(x)$ is equivalent to minimizing $\varphi(x)^2$. Notice

$$\varphi(x)^2 = \left( 1 + \frac4x \right) + ( 1 - x ) - 2\sqrt{\left( 1 + \frac4x \right)(1-x)} = 2 + \frac4x - x - 2\sqrt{\frac4x - x - 3}$$ If we set $\psi(x) = \sqrt{\frac4x - x - 3}$, we have

$$\varphi(x)^2 = 2 + (\psi(x)^2 + 3) - 2\psi(x) = 4 + (\psi(x)-1)^2 \ge 4\implies \varphi(x) \ge 2$$

In order for what's on the right to become an equality, we need $$\psi(x) = 1 \iff \frac4x - x - 3 = 1^2 \iff x(x+4) = 4 \implies x = \pm\sqrt{8} - 2$$

Since $x_+ = \sqrt{8}-2 \approx 0.8284 \in (0,1]$, we have

$$\psi(x_+) = 1 \quad\implies\quad \varphi(x_+) = 2 \quad\implies\quad \min_{x \in (0,1]} \varphi(x) = \varphi(x_+) = 2$$

Update

For something more advanced, the inequality $\varphi(x) \ge 2$ can be proved using Aczél's inequality.

Given real numbers $a_0, b_0 \in \mathbb{R}$ and vectors $\vec{a}, \vec{b} \in \mathbb{R}^n$ such that $a_0^2 > |\vec{a}|^2$, $b_0^2 > |\vec{b}|^2$, we have $$(a_0b_0 - \vec{a} \cdot \vec{b})^2 \ge (a_0^2 - |\vec{a}|^2)(b_0^2 - |\vec{b}|^2)$$ with equality if and only if $\displaystyle\;\frac{\vec{a}}{a_0} = \frac{\vec{b}}{b_0}$.

For the problem at hand, let $u = \sqrt{1 + \frac{4}{x}}$ and $v = \sqrt{1 - x}$. Notice for $x \in (0,1]$, we have $$u^2 - 1 = \frac{4}{x} > 0\quad\text{ and }\quad 1 - v^2 = 1 - (1-x) = x > 0$$ Using Aczél's inequality, we obtain $$ \varphi(x) = u-v = u\cdot 1 - 1 \cdot v \ge \sqrt{(u^2-1)(1-v^2)} = \sqrt{\frac4x \cdot x} = \sqrt{4} = 2 $$ with equality when $\displaystyle\;\frac{1}{u} = \frac{v}{1} \iff uv = 1$. The rest of argument is same as above and I won't repeat them here.