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⁡(2sin⁡x)>12x(π−x)−572π2,∀x∈(π6,π2)\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\tan x>x for 0<x<π20< x <\dfrac{\pi}{2}. Now if we substitute xx with π2−x\dfrac{\pi}{2}-x we'll obtain cot⁡x>π2−x\cot x>\dfrac{\pi}{2}-x. Now integrating our inequality we obtain ∫π6xcot⁡t dt>∫π6x(π2−t) dt.\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⁡12>(πx2−x22)−(π2⋅π6−12(π26))⇔\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 ⇔ln⁡(2sin⁡x)>12x(π−x)−572π2.\Leftrightarrow \ln(2\sin x)>\frac{1}{2}x(\pi-x)-\frac{5}{72}\pi^2.

Problem 2. For 0<x<π20< x <\dfrac{\pi}{2} prove the following inequality sin⁡x≤x(π−x)2.\sin x \leq \frac{x(\pi-x)}{2}.

Problem 3. For b≥1b\geq 1 prove the following inequality ab≤ea+b(ln⁡b−1).ab\leq e^a+b(\ln b -1).

#Calculus #Inequalities #Integration #Goldbach'sConjurersGroup

Note by Nicolae Sapoval
7 years, 5 months ago

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Comments

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

Daniel Liu - 7 years, 5 months ago

Very nice problems keep it up nicolae

abdulmuttalib lokhandwala - 7 years, 1 month ago

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

Dinesh Chavan - 7 years, 3 months ago

Log in to reply

I could only solve it by considering two cases. Case 1. If a>x>ln⁡b≥0a>x> \ln b \geq 0 then in this interval ex≥b≥1e^x \geq b \geq 1. Thus ∫ln⁡baexdx≥∫ln⁡babdx\displaystyle \int_{\ln b}^{a} e^x \mathrm{d}x \geq \displaystyle \int_{\ln b}^{a} b \mathrm{d}x. On expanding we have ea−b≥b(a−ln⁡b)⇒ab≤ea+b(ln⁡b−1)e^a-b \geq b(a-\ln b) \Rightarrow ab \leq e^a+ b(\ln b -1).  \space Case 2 If ln⁡b≥x≥0>a\ln b \geq x \geq 0 >a then in this interval b≥ex b\geq e^x. Thus ∫aln⁡bbdx≥∫aln⁡bexdx\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)≥b−ea⇒ab≥ea+b(ln⁡b−1)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 - 7 years, 2 months ago

thanks! :)

hemang sarkar - 6 years, 11 months ago
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