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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

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    $\begingroup$ Try squaring both sides. $\endgroup$ – David Apr 27 '14 at 6:13
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Since $x^x$ is positive, you can square both sides without worrying about the inequality. Then replace $x^x = n$, so the inequality becomes $n^2 \ge 2n - 1$. Rearrange, $n^2 - 2n + 1\ge 0$. Factor, $(n-1)^2 \ge 0$, the trivial inequality.

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  • $\begingroup$ @Cole the question specifies $x \geq 0$ so it can't be -1. $\endgroup$ – David Z Apr 28 '14 at 1:24
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By AM-GM: $$a=\frac{(2a-1)+1}2\geq\sqrt{2a-1}$$

Taking $a=x^x$, since $2x^x-1$ is positive:

$$x^x\geq \sqrt{2x^x-1}$$

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Let $y=x^x \ge 0$. Then you have $y \ge \sqrt{2y-1} \iff y^2 \ge 2y - 1 \iff (y-1)^2 \ge 0$.

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Note that $a^2-2a+1=(a-1)^2\geq0$ for all real number $a$. Then, for $a>0$ we have $a^2\ge2a-1$ and $a\ge\sqrt{2a-1}$. Making $a=x^x$ for $x>0$ we obtain $x^x\ge\sqrt{2x^x-1}$.

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Taking a derivative (or just observing) shows that the left side grows quicker than the right. Thus we just need to establish that the LHS is greater than the right at $x=0$. We see that both sides are equal. Thus since the left grows faster, the inequality holds.

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