For $x \ge 1$, is $2^{\sqrt{x}} \ge x$ For $x \ge 1$, is $2^{\sqrt{x}} \ge x$
The answer appears to me to be yes since:
$$\sqrt{x} \ge \log_2 x$$
Here's my reasoning:
(1) For $x > 16$, $5\sqrt{x} > 4\sqrt{x} + 4$
(2) For $x > 25$, $x > 5\sqrt{x}$
(3) So, $x > 25$, $2x > x + 4\sqrt{x} + 4$ and it follows that $\sqrt{2x} > \sqrt{x} +2$
(4) So, as $x$ doubles, $\sqrt{2x} > \sqrt{x} + 2$ but $\log_2(2x) = \log_2(x) +1$
Is my reasoning correct?  Is there a simpler way to analyze this?

The claim is in fact false. Take $x=9$, we have 
$$2^{\sqrt{9}}=2^3=8<9.$$
