Integration, just as derivation, reveals a new approach to proving the inequalities. Let's take a detailed view on inequalities solved by or involving intgrals.

**Theorem 1.** If for all $x\in[a,b]$ the following inequality holds
$f(x)\geq g(x),$
then for all $x\in[a,b]$ we have
$\int^x_af(t)\,dt\geq\int^x_ag(t)\,dt.$

I won't present a formal proof, but rather just a simple image to ilustrate the idea.

Now armed with this theorem let's solve few very basic problems.

**Problem 1.** Prove the following inequality
$\ln(2\sin x)>\frac{1}{2}x(\pi-x)-\frac{5}{72}\pi^2,\forall x\in\left(\frac{\pi}{6},\frac{\pi}{2}\right)$

*Solution.* We know the very basic inequality $\tan x>x$ for $0< x <\dfrac{\pi}{2}$. Now if we substitute $x$ with $\dfrac{\pi}{2}-x$ we'll obtain $\cot x>\dfrac{\pi}{2}-x$. Now integrating our inequality we obtain
$\int^x_{\frac{\pi}{6}}\cot t\,dt>\int^x_{\frac{\pi}{6}}\left(\dfrac{\pi}{2}-t\right)\,dt.$
Calculating we get
$\ln(\sin x)-\ln\frac{1}{2}>\left(\frac{\pi x}{2}-\frac{x^2}{2}\right)-\left(\frac{\pi}{2}\cdot\frac{\pi}{6}-\frac{1}{2}\left(\frac{\pi^2}{6}\right)\right)\Leftrightarrow$
$\Leftrightarrow \ln(2\sin x)>\frac{1}{2}x(\pi-x)-\frac{5}{72}\pi^2.$

**Problem 2.** For $0< x <\dfrac{\pi}{2}$ prove the following inequality
$\sin x \leq \frac{x(\pi-x)}{2}.$

**Problem 3.** For $b\geq 1$ prove the following inequality
$ab\leq e^a+b(\ln b -1).$

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## Comments

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TopNewestThis is actually a really interesting topic that I have never heard of! Thanks for sharing!

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Very nice problems keep it up nicolae

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Sir, Can You give Solution to Problem 3,,I am not able to do it. I have done With 2nd one

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I could only solve it by considering two cases.

Case 1. If $a>x> \ln b \geq 0$ then in this interval $e^x \geq b \geq 1$. Thus $\displaystyle \int_{\ln b}^{a} e^x \mathrm{d}x \geq \displaystyle \int_{\ln b}^{a} b \mathrm{d}x$. On expanding we have $e^a-b \geq b(a-\ln b) \Rightarrow ab \leq e^a+ b(\ln b -1)$. $\space$Case 2If $\ln b \geq x \geq 0 >a$ then in this interval $b\geq e^x$. Thus $\displaystyle \int_{a}^{\ln b} b\mathrm{d}x \geq \displaystyle \int_{a}^{\ln b} e^x \mathrm{d}x$. Which on expanding gives us $b(\ln b -a) \geq b- e^a \Rightarrow ab \geq e^a + b(\ln b -1)$. PS: Let me know if you find a mistake or an easier way, all criticisms are welcomed.Log in to reply

thanks! :)

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