TITLE: Very tough inequality I cannot prove
QUESTION [10 upvotes]: Suppose $x \geq 0 $ is real number, then
$$ x^x \geq \sqrt{ 2x^x - 1 } $$
How can I show this? Also, what is the greometrical implication of such inequality ? thanks

REPLY [8 votes]: By AM-GM: 
$$a=\frac{(2a-1)+1}2\geq\sqrt{2a-1}$$
Taking $a=x^x$, since $2x^x-1$ is positive:
$$x^x\geq \sqrt{2x^x-1}$$