How to solve this irrational inequality with different signs?

Question

I have the following inequality:

$$x\geq-\frac{1}{2}\sqrt{3x^2+4}$$

My first impulse is to square it, but firstly, $x$ can have any sign. And secondly, the right part is negative.

I have no idea what to do, please help. Or at the least give me a hint.

Answer 1

Yes squaring is correct! Note that the inequality clearly holds when $x\ge0$, so consider when $x=-k$, where $k$ is a positive real number. Then

$$x\ge-0.5\sqrt{3x^2+4}\implies k\le0.5\sqrt{3k^2+4}\implies k^2\le0.25(3k^2+4)=0.75k^2+1...$$ I'm sure you can go on.

Answer 2

If $$x\geq 0$$ then our inequality is true. For $$x<0$$ we get: $$-x\le \frac{1}{2}\sqrt{3x^2+4}$$ Now all is positive and we get by squaring $$4x^2\le 3x^2+4$$ $$x^2\le 4$$ So?