# Inequality involving multiple square roots

Wolfram alpha solves $\sqrt{x+1}\ge\sqrt{x+2}+\sqrt{x+3}$ for $x$, and answers $x=-2/3(3+\sqrt{3})$. How did it do it? Thanks!

• Are you sure about the inequality? The inequality is trivially false for $x \geq -1$ and undefined for $x<-1$, since you cannot compare complex numbers. – mlk Jun 13 '14 at 18:04
• Out of curiosity, I asked Wolfram Alpha to solve the inequality. To my disappointment, it said there is no solution. – André Nicolas Jun 13 '14 at 18:13

We need $x\ge -1$ in order for all roots to be defined. Then the right hand side is positive, hence in lets divide by the positive number $\sqrt{x+1}$ (and note that $x+1>0$): $$1\ge\sqrt{1+\frac1{1+x}}+\sqrt {1+\frac{2}{1+x}}\ge 2$$ contradiction!
I'd say it's okay. That the number $X=-2/3(3+\sqrt{3})$ is the solution of the equation $\sqrt{x+1} >\color{red}{=} \sqrt{x+2}+\sqrt{x+3}:$
$\sqrt{X+1}=i\sqrt{1/3(3+2\sqrt{3})}$
$\sqrt{X+2}+\sqrt{X+3}=i\sqrt{1/3 (3+2\sqrt{3})}$