Inequality with too many square roots

Question

I've been trying to prove the following inequality but the square roots really give me a hard time. $$x\leq \sqrt{1-\left(\frac{x^2+2}{4x}\right)^2}+\sqrt{1-\left(\frac{-x}{2(\sqrt{-2(x^2-4)}-3)}\right)^2}$$ for $x\in[1;\sqrt2]$. I found that both sides are equal at $x=\sqrt2$.

Answer 1

Let $$A := \frac{x^2 + 2}{4x}$$ and $$B := \frac{-x}{2(\sqrt{-2(x^2-4)}-3)} = \frac{x}{6 - 2\sqrt{8 - 2x^2}} = \frac{x(6 + 2\sqrt{8 - 2x^2})}{8x^2 + 4}.$$

We have $$\sqrt{8 - 2x^2} = \frac{(8-2x^2) - 2^2}{\sqrt{8-2x^2} + 2} + 2 \le \frac{4 - 2x^2}{2 + 2} + 2 = 3 - x^2/2$$ and $$B \le \frac{x[6 + 2(3 - x^2/2)]}{8x^2 + 4} = \frac{x(12 - x^2)}{8x^2 + 4} =: C.$$

It suffices to prove that $$x \le \sqrt{1 - A^2} + \sqrt{1 - C^2}$$ or $$x/2 - \sqrt{1 - C^2} \le \sqrt{1 - A^2} - x/2$$ or $$\frac{(x/2)^2 - (1 - C^2)}{x/2 + \sqrt{1 - C^2}} \le \frac{1 - A^2 - (x/2)^2}{\sqrt{1 - A^2} + x/2}.$$

We have $$1 - A^2 - (x/2)^2 = \frac{(2 - x^2)(5x^2 - 2)}{16x^2} \ge 0 \tag{1}$$ and $$(x/2)^2 - (1 - C^2) = \frac{(2-x^2)^2(17x^2-4)}{16(2x^2+1)^2} \ge 0. \tag{2}$$

It is easy to prove that $C \le \frac{11}{12}$. Thus, we have $$\sqrt{1 - C^2} \ge \sqrt{1 - \frac{121}{144}} > \frac{\sqrt 2}{5} \ge \frac{x}{5}. \tag{3}$$

Using AM-GM, we have $A \ge \frac{2\sqrt{x^2 \cdot 2}}{4x} = \frac{\sqrt 2}{2}$. Thus, we have $$\sqrt{1 - A^2} \le \sqrt{1 - 1/2} < 3/4 \le 3x/4. \tag{4}$$

Using (1)-(4), it suffices to prove that $$\frac{1}{x/2 + x/5}\cdot \frac{(2-x^2)^2(17x^2-4)}{16(2x^2+1)^2} \le \frac{1}{3x/4 + x/2}\cdot \frac{(2 - x^2)(5x^2 - 2)}{16x^2}.$$

Using $2x^2 + 1 - 3x = (2x - 1)(x - 1) \ge 0$, it suffices to prove that $$\frac{1}{x/2 + x/5}\cdot \frac{(2-x^2)^2(17x^2-4)}{16\cdot (3x)^2} \le \frac{1}{3x/4 + x/2}\cdot \frac{(2 - x^2)(5x^2 - 2)}{16x^2}$$ or $$25(2 - x^2)(17x^2 - 4) \le 126(5x^2 - 2)$$ which is true. (Hint: Let $y = x^2 \in [1, 2]$.)

We are done.