Show that for all $x \ge 1$, $\log(x) \le \sqrt{x} - \frac{1}{\sqrt{x}}$

Question

This is problem #9 from section 3.5 of Buck's Advanced Calculus. The section is about Taylor's Theorem so my instinct was to try to find some appropriate function with a Taylor polynomial that could relate the derivatives of $\log(x)$ to the right side of the inequality, but I can't seem to manage it. I also was trying to think of the right side as $x^{1/2}-x^{-1/2}$ and see what could come of that, but again hit a roadblock. Any advice on how to continue?

Answer 1

By setting $x = \mathrm{e}^{2t}$, the inequality becomes equivalent to $t \leq \sinh t$ for $t \geq 0$, and this is readily apparent from the Maclaurin series of $\sinh t$.

Answer 2

Hint. Define $f(x)=\sqrt{x}-\frac{1}{\sqrt{x}}-\log(x)$. Show that (a) $f(1)=0$ (that's easy!), and (b) $f'(x)\geq 0$ for $x\geq 1$. The inequality you want to prove follows from (a) and (b).

Answer 3

Hint

Consider that you look for the minimum value of function $$ f(x)= \sqrt{x} - \frac{1}{\sqrt{x}}-\log(x)$$ Since $x>0$, let $x=t^2$ and consider $$g(t)=t-\frac 1t-2\log(t)$$ The first derivative $$g'(t)=\frac{(t-1)^2}{t^2}$$ cancels at $t=1$