Inequality $\log x\le \frac{2}{e} \, \sqrt{x}$

Question

The inequality $$\log x \le \frac{2}{e} \, \sqrt{x},$$ where $\log x$ denotes the natural logarithm, is used in the proof of Theorem 4.7 in Apostol's Analytic Number Theory.

It seems that the inequality is not very difficult to prove using calculus. We could simply find maximum/minimum of some function like $f(x)= \frac2e \sqrt{x} - \log x$ or $g(x)=\frac{\log x}{\sqrt x}$.

Are there some other methods how this inequality can be proved? Is there a way in which this inequality can be seen more directly, without having to calculate critical points of some auxiliary function?

Answer 1

With the substitution $x = e^{2(u+1)}$ the inequality $$ \log x \le \frac{2}{e} \, \sqrt{x} $$ becomes $$ e^u \ge 1 + u \tag{*} $$ which is a well-known estimate for the exponential function. Equality holds if and only if $u = 0$, corresponding to $x = e^2$ in the original inequality.

$(*)$ is trivial for $u \le -1$ and can for example be shown using the Taylor series for $u > -1$. It also follows – as Jack said in a comment – from the convexity of the exponential function: the graph lies above the tangent line at $u = 0$.

(This approach was inspired by Jack D'Aurizio's answer.)

Answer 2

With a change of variable, $x=e^{2t}$, the given inequality is equivalent to: $$ t e^{-t} \leq \frac{1}{e}. \tag{1}$$ Let we set $f(t)=t e^{-t}$. We have $f'(t)=(t-1)e^{-t}$, hence $t=1$ is the only stationary point of $f(t)$. Quite trivially, it is an absolute maximum, hence: $$ f(t) \leq f(1) = \frac{1}{e} \tag{2}$$ follows.

Answer 3

Letting $x=(eu)^2$, the inequality is equivalent to

$$\log u\le u-1$$