<s>[INST] 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?
 [/INST] The inequality doesn't always hold, as can be seen by taking $x=9$.
Notice that your question is about comparing $a^b$ and $b^a$ with $a=2,b=\sqrt{x}$. It is well known that $a^b \leq b^a$ if and only if $$a^{1/a} \leq b^{1/b} $$
which in our case is
$$\sqrt{2} \leq b^{1/b}.$$
This inequality holds iff $$2 \leq b \leq 4 $$
which translates to $$ 4 \leq x \leq 16 .$$
Thus your inequality holds for $x \in [1,4] \cup [16,\infty)$.
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