For $x \ge 1$, is $2^{\sqrt{x}} \ge x$

Question

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?

Answer 1

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 $x \in [1,4] \cup [16,\infty)$.

Answer 2

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

Answer 3

Instead of considering $$f(x)=2^{\sqrt x}-x$$ consider function and derivatives $$g(x)={\sqrt x}\log 2-\log(x)$$ $$g'(x)=\frac{\log (2)}{2 \sqrt{x}}-\frac{1}{x}$$ $$g''(x)=\frac{1}{x^2}-\frac{\log (2)}{4 x^{3/2}}$$ The first derivative cancels at $x_0=\frac{4}{\log ^2(2)}$ and, for this value $$g(x_0)=2-2\log (2)+2 \log (\log (2))\approx -0.11932\qquad g''(x_0)=\frac{\log ^4(2)}{32}$$ So, $x_0$ coressponds to a unique minimum. Since, when $x\to 0$, $g(x)\to \infty$, then there are two solutions $x_1,x_2$ to $g(x)=0$; for any $x_1 < x < x_2$, $g(x) <0$ and then $f(x) >0$ if $0<x<x_1$ and $x_2 <x <\infty$.