Prove or disprove that : $x^{\frac{\sin\left(x\right)}{x}}> \sin\left(x\right)+\frac{1}{x-1}$ for $x\geq \pi$

Question

Hi and sorry for the inconvenience of my last question .

I work again with the function :

$$f(x)=x^{\frac{\sin\left(x\right)}{x}}$$

Working again with the software Desmos I found that :

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Claim:

Let $x\geq \pi$ then we have :

$$f(x)> \sin\left(x\right)+\frac{1}{x-1}$$

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I cannot show it but I can prove a weaker result easily :

Let $x\geq \pi$ then we have :

$$f(x)>\sin(x)$$

The proof is really basic just taking the logarithm we need to show for $\sin(x)>0$:

$$\frac{\ln\left(\sin\left(x\right)\right)}{\sin\left(x\right)}<\frac{\ln\left(x\right)}{x}$$

Wich is obvious because we have :

$$\frac{\ln\left(\sin\left(x\right)\right)}{\sin\left(x\right)}\leq 0<\frac{\ln\left(x\right)}{x}$$

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I find this problem interesting because it evaluates some extrema of $f(x)$ wich we are unable to find explicitly .

Question :

How to prove or disprove the claim?

Thanks for your try and your efforts in this sense .

Answer 1

You may use the well-known inequality $e^x\ge 1+x$, therefore:

$$x^\frac{\sin(x)}{x}=e^{\ln(x)\frac{\sin(x)}{x}}\ge 1+\ln(x)\frac{\sin(x)}{x}$$

So it suffices to prove that:

$$1+\ln(x)\frac{\sin(x)}{x}>\sin(x)+\frac{1}{x-1}$$

Or:

$$\sin(x)\left(1-\frac{\ln(x)}{x}\right)<1-\frac{1}{x-1}$$

Now if $\pi\le x\le2\pi$ you have LHS$\,\le 0$ and RHS$\,>0$ and you are done.

If $x>2\pi$ you may ignore $\sin(x)$ and directly prove that:

$$1-\frac{\ln(x)}{x}<1-\frac{1}{x-1}$$

Or:

$$\ln(x)>\frac{x}{x-1}$$

The last inequality is obvious ($\ln$ is increasing, $\frac{x}{x-1}$ is decreasing, so $\ln(x)>\ln(2\pi)>\frac{2\pi}{2\pi-1}>\frac{x}{x-1}$)

Note that $\ln(x)>\frac{x}{x-1}$ is not true for $1\le x\le 3.85\ldots$, that's why $\pi\le x\le 2\pi$ had to be proved separately.

Answer 2

This is not a proof.

Considering the function $$f(x)=x^{\frac{\sin\left(x\right)}{x}}- \sin\left(x\right)-\frac{1}{x-1}$$ its $n^{\text{th}}$ minimum is extremely close to $t=(4n+1)\frac \pi 2$ (in fact, $\color{red}{\text{just above}}$).

Using a series expansion around this point, we have $$f^{\text{min}}_n=\Bigg[t^{\frac{1}{t}}-\frac{t}{t-1}\Bigg]+\Bigg[\frac{1}{(t-1)^2}-t^{\frac{1}{t}-2} \log \left(\frac{t}{e}\right)\Bigg](x-t)+O((x-t)^2)$$

For any $n \geq 1$, the first coefficient is always positive, $$t^{\frac{1}{t}}-\frac{t}{t-1}= \sum_{n=1}^\infty \left(\frac{\log ^n(t)}{n!}-1\right)\,t^{-n}$$

the second one always negative but since $t >x$ all of that is positive.

Comparing the results of the constant term of this approximation with those from a full optimization $$\left( \begin{array}{ccc} n & \text{approximation} & \text{solution} \\ 1 & 0.1541683467 & 0.1541674660 \\ 2 & 0.1299472506 & 0.1299360987 \\ 3 & 0.1076980424 & 0.1076927780 \\ 4 & 0.0919903207 & 0.0919877355 \\ 5 & 0.0805424747 & 0.0805410723 \\ 10 & 0.0510386080 & 0.0510384367 \\ 15 & 0.0382209486 & 0.0382209025 \\ 20 & 0.0309002613 & 0.0309002437 \\ 25 & 0.0261085589 & 0.0261085507 \\ 50 & 0.0152170441 & 0.0152170433 \\ 100 & 0.0086952495 & 0.0086952494 \\ 1000 & 0.0012334307 & 0.0012334307 \\ 10000 & 0.0001599340 & 0.0001599340 \end{array} \right)$$