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# Playing with Integrals: Inequalities and Integrals 1

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

Note by Nicolae Sapoval
3 years, 2 months ago

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This is actually a really interesting topic that I have never heard of! Thanks for sharing! · 3 years, 2 months ago

Very nice problems keep it up nicolae · 2 years, 10 months ago

thanks! :) · 2 years, 8 months ago

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

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 2 If $$\ln b \geq x \geq 0 >a$$ 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. · 3 years ago