# $G(x)\sqrt{A(x)}\geq \lambda \, \, \, \, \, \lambda \in \mathbb{R}$ e.g $2x\sqrt{1-x^{2}}\geq-1$

I realized I do not know how to solve a particular type of inequalities.

I needed to solve:

$$2x\sqrt{1-x^{2}}\geq-1$$ Usually with the irrational inequalities where the variable is just below the root and in the other member of the equation I set the following systems: $$\sqrt{A(x)}\geq B(x)$$ $$\left\{\begin{matrix} A(x)\geq 0\\ B(x)\geq 0\\ A(x)^{2}\geq B(x)^{2} \end{matrix}\right.$$ $$\cup$$

$$\left\{\begin{matrix} A(x)\geq 0\\ B(x)\leq 0\\ \end{matrix}\right.$$

But in this case we do not have the variable in the second term and being an inequality I remember I can not to divide by a variable. I did not find on my book a clear analysis of these cases.

I am interested in the following cases: $$G(x)\sqrt{A(x)}\geq \lambda \, \, \, \, \, \lambda \in \mathbb{R}$$ $$G(x)\sqrt{A(x)}\geq B(x)$$

Thank you a lot for your help!!

$$2x\sqrt{1-x^{2}}\ge 1\implies x>0, 1-x^2>0.$$ So, squaring we have
$$4x^2(1-x^2)\ge 1,$$ or
$$(2x^2-1)^2=4x^4-4x^2+1\le 0.$$ The only solution is $x=\frac{1}{\sqrt 2}$ (remember, $x>0.$)