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
\[ab\leq e^a+b(\ln b -1).\]

## Comments

Sort by:

TopNewestThis is actually a really interesting topic that I have never heard of! Thanks for sharing! – Daniel Liu · 3 years ago

Log in to reply

Very nice problems keep it up nicolae – Abdulmuttalib Lokhandwala · 2 years, 8 months ago

Log in to reply

thanks! :) – Hemang Sarkar · 2 years, 6 months ago

Log in to reply

Sir, Can You give Solution to Problem 3,,I am not able to do it. I have done With 2nd one – Dinesh Chavan · 2 years, 10 months ago

Log in to reply

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. – Abhijeeth Babu · 2 years, 10 months agoLog in to reply