Prove for all $x\geq 1$, $\log x \leq \sqrt{x}-\frac{1}{\sqrt{x}}$. What I tried:
Since for $t>0$, $\frac{1}{t}<\frac{1}{2t}(\sqrt{t}+\frac{1}{\sqrt{t}})$, then let $x\geq 1$, and integral both sides from $1$ to $x$, then can get the title.
But this question has a hint say first show $\frac{1}{x}<\frac{1}{2}(\frac{1}{x^{1+\delta}}+\frac{1}{x^{1-\delta}})$, for any $\delta, 0<\delta<1$. I don't know how to prove the hint.
And since it's a exercise after Taylor series, can it be proved use Taylor series directly?
Thanks!
 A: Another way...
We will prove:
$$
\log x \leq \sqrt{x}-\frac{1}{\sqrt{x}},\qquad x \ge 1
$$
let $x = e^t$:
$$
t \le e^{t/2}-e^{-t/2},\qquad t \ge 0
\\
\frac{t}{2} \le \sinh\frac{t}{2},\qquad t \ge 0
$$
it is enough to prove
$$
u \le \sinh u,\qquad u \ge 0
$$
This is clear because the Maclaurin series of $\sinh u$ is $u$ plus nonnegative terms and converges for all $u$:
$$
\sinh u  = \sum_{k=0}^\infty \frac{u^{2k+1}}{(2k+1)!}
$$
A: METHODOLOGY $1$:  Using Taylor's Theorem
First we let $y=\sqrt x$.  Then, the inequality $\log(x)\le \sqrt x-\frac1{\sqrt x}$ for $x\ge 1$ is equivalent to the inequality 
$$y\log(y)\le \frac12\left(y^2-1\right)$$
for $y\ge 1$.
Using Taylor's Theorem (with remainder) for $\log(y)$ we see that $\log(y)\le y-1+\frac12(y-1)^2$ for $y\ge 1$.  Hence, we have for $y\ge 1$
$$\begin{align}
y\log(y)&=(y-1)\log(y)+\log(y)\\\\
&\le (y-1)^2-\frac12(y-1)^3+(y-1)-\frac12 (y-1)^2\\\\
&=(y-1)+\frac12(y-1)^2\\\\
&= \frac12 (y^2-1)\end{align}$$
And we are done!

METHODOLOGY $2$:  Using the Mean Value Theorem
Let $f(x)=\log(x)-\sqrt{x}+\frac1{\sqrt x}$.  Note that $f(1)=0$ and for $x\ge 1$
$$f'(x)=-\frac{(\sqrt x-1)^2}{2x^{3/2}}\le 0$$
Can you finish?

A: Using the fact that 
$x\ge 1 \iff \sqrt{x}\ge 1$, and
$$\ln(x)=2\ln(\sqrt{x}),$$
it is equivalent to prove that
$$2\ln(x)\le x-\frac{1}{x}$$
and with $f(x)=2\ln(x)-x+\frac 1x$, 
$$f'(x)=\frac 2x-1-\frac{1}{x^2}=-\Bigl(\frac{x-1}{x}\Bigr)^2$$
$ f$ is decreasing at $[1,+\infty)$ and $ f(1)=0$ implies that for $x\ge 1$
$$f(x)\le f(1) \iff 2\ln(x)\le x-\frac 1x$$
$$\iff \ln(x)\le \sqrt{x}-\frac{1}{\sqrt{x}}$$
A: Following the hint, try to prove $$x^{\delta}+x^{-\delta}\geq2$$
for all $x\geq1$ and $0<\delta<1$. Substituting $y:=x^{\delta}$ the problem reduces to minimizing 
$$g(y)=y+\frac{1}{y}$$
where $y=x^{\delta}\geq 1$. If you can prove $g(y)\geq2$ then this implies that the function 
$$f(x)=\sqrt{x}-\frac{1}{\sqrt x}-\log(x)$$
is increasing when $x\geq 1$. Hint:
$$f'(x)=\frac{1}{2}\left(\frac{1}{x^{3/2}}+\frac{1}{x^{1/2}}\right)-\frac{1}{x}.$$
Can you use the previous result here?
A: Let $f(x)=\sqrt x- 1/\sqrt x -\log x.$ For all $x>1$ we have $$f'(x)=1/(2\sqrt x) + 1/(2x\sqrt x) -1/x =(2x\sqrt x)^{-1}\cdot(-1+\sqrt x)^2>0.$$ 
A: Writing $x=u^2$ with $u\ge1$, the inequality to prove becomes
$$2\log u\le u-{1\over u}$$
Now
$$2\log u=\int_1^u{dt\over t}+\int_1^u{dt\over t}=\int_1^u{dt\over t}+\int_1^u{dt\over u+1-t}=\int_1^u{(u+1)dt\over t(u+1-t)}$$
and
$$1\le t\le u\implies (t-1)(t-u)\le0\implies u\le t(u+1-t)$$
It follows that
$$2\log u=(u+1)\int_1^u{dt\over t(u+1-t)}\le(u+1)\int_1^u{dt\over u}={(u+1)(u-1)\over u}={u^2-1\over u}=u-{1\over u}$$
