Prove for all $x\geq 1$, $\log x \leq \sqrt{x}-\frac{1}{\sqrt{x}}$.

Question

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!

Answer 1

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 $u$:

$$ \sinh u = \sum_{k=0}^\infty \frac{u^{2k+1}}{(2k+1)!} $$

Answer 2

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!

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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?

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Answer 3

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 $x\ge 1$

$$f(x)\le f(1) \iff 2\ln(x)\le x-\frac 1x$$ $$\iff \ln(x)\le \sqrt{x}-\frac{1}{\sqrt{x}}$$

Answer 4

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?

Answer 5

Let $f(x)=\sqrt x- 1/\sqrt x -\log x.$ For all $x>1$ we have $$f'(x)=1/(2\sqrt x) + 1/(2x\sqrt x) -1/x =(2x\sqrt x)^{-1}\cdot(-1+\sqrt x)^2>0.$$

Answer 6

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

$$2\log u=(u+1)\int_1^u{dt\over t(u+1-t)}\le(u+1)\int_1^u{dt\over u}={(u+1)(u-1)\over u}={u^2-1\over u}=u-{1\over u}$$