Solve the inequality $\sqrt{7+2016x}+49\le 4x^2+\sqrt{2018x}$

Question

Solve the inequality $$\sqrt{7+2016x}+49\le 4x^2+\sqrt{2018x}$$

My work

1) I used that $A\le B$ and $A,B>0$ $\Leftrightarrow A^2\le B^2$

2) I know the answer [wolframalpha]1

3) I multyplied both sides $\sqrt{7+2016x}-49$

Is there any simple way how to solve this inequality?

Answer 1

Yes, of course!

It's $$4x^2-49+\sqrt{2018x}-\sqrt{7+2016x}\geq0$$ or $$(2x-7)\left(2x+7+\frac{1}{\sqrt{2018x}+\sqrt{7+2016x}}\right)\geq0$$ or $$x\geq\frac{7}{2}.$$

Answer 2

If $x \geq 3.5$, $4x^2 \geq 49$ and $7+2016x \leq 2018x$, so the inequality holds.

If $x < 3.5$, then $4x^2 < 49$ and $7+2016x > 2018x$, so the inequality doesn’t hold.

In other words: the inequality holds iff $x \geq 3.5$.

Answer 3

Note that the expression is defined only for non-negative $x$. Now, $$\sqrt{7+2016x}+49\le 4x^2+\sqrt{2018x}$$ $$\iff 4x^2-49 + \sqrt{2018x}-\sqrt{2016x+7}\ge 0$$ $$\iff (2x-7)(2x+7)+\dfrac{2x-7}{\sqrt{2018x}+\sqrt{2016x+7}}\ge 0 $$ $$\iff 2x-7\ge 0$$