# Prove/disprove that $a \le b$ where $4x \le y$, $a = \sqrt{-8x + 2\sqrt x + 1}$, $b = \sqrt[3]{-48y + 3b\sqrt[3] y + 1}$ and $a, b, x, y \ge 1$.

Given that $$a, b\ge 1$$ and $$x, y > 0$$ satisfied $$\large 2\sqrt[4]{x} \le a = \sqrt{-8x + 2\sqrt x + 1}$$ and $$\large 2\sqrt[6]{y} \le b = \sqrt[3]{-48y + 3b\sqrt[3] y + 1}$$, prove/disprove that if $$4x \le y$$ then $$a \le b$$.

This is the weirdest problem that I've ever seen in my lifetime. I wish I could pour my efforts into this problem but I just can't.

If the inequality is not correct, what minor change could have been done to the inequality so that it is correct for $$\forall a, b \ge 1$$ and $$x, y > 0$$?

• What are the steps that you have already taken, and why have they failed? It seems to me that the first thing to do would be to check which values are allowed for x, b, and y, since they are under a root. Have you done this? E.g. when is $-8x+2\sqrt{x}+1 \geq 0$? – PoorYorick Jun 21 at 17:14
• The equation for $y$ is indeed a little weird, the $3b\sqrt[3]y$ term is suggesting something although I can't see what exactly it is yet. – user10354138 Jun 21 at 17:24

Hint: It must be $$-8x+2\sqrt{x}+1\geq 0$$, this is only fulfilled for $$0\le x\le \frac{1}{4}$$ so your inequality is not true.

• However, then the question is terribly phrased. "Given A, prove B" necessitates that A is true. Otherwise the question does not even have to be attempted. It can be left blank, since the condition for solving it is not fulfilled. – PoorYorick Jun 21 at 17:40
• I have fixed the question. Please double-check the new problem. There were some misinterpretations while I was typing it. – Lê Thành Đạt Jun 22 at 0:17