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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
2 years, 9 months ago

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This is actually a really interesting topic that I have never heard of! Thanks for sharing! Daniel Liu · 2 years, 9 months ago

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Very nice problems keep it up nicolae Abdulmuttalib Lokhandwala · 2 years, 5 months ago

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thanks! :) Hemang Sarkar · 2 years, 3 months ago

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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 Dinesh Chavan · 2 years, 7 months ago

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@Dinesh Chavan 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. Abhijeeth Babu · 2 years, 7 months ago

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