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[a,b]x\in[a,b] the following inequality holds f(x)g(x),f(x)\geq g(x), then for all x[a,b]x\in[a,b] we have axf(t)dtaxg(t)dt.\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(2sinx)>12x(πx)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 tanx>x\tan x>x for 0<x<π20< x <\dfrac{\pi}{2}. Now if we substitute xx with π2x\dfrac{\pi}{2}-x we'll obtain cotx>π2x\cot x>\dfrac{\pi}{2}-x. Now integrating our inequality we obtain π6xcottdt>π6x(π2t)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(sinx)ln12>(πx2x22)(π2π612(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(2sinx)>12x(πx)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 sinxx(πx)2.\sin x \leq \frac{x(\pi-x)}{2}.

Problem 3. For b1b\geq 1 prove the following inequality abea+b(lnb1).ab\leq e^a+b(\ln b -1).

Note by Nicolae Sapoval
5 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 - 5 years, 9 months ago

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Very nice problems keep it up nicolae

abdulmuttalib lokhandwala - 5 years, 5 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 - 5 years, 7 months ago

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I could only solve it by considering two cases. Case 1. If a>x>lnb0a>x> \ln b \geq 0 then in this interval exb1e^x \geq b \geq 1. Thus lnbaexdxlnbabdx\displaystyle \int_{\ln b}^{a} e^x \mathrm{d}x \geq \displaystyle \int_{\ln b}^{a} b \mathrm{d}x. On expanding we have eabb(alnb)abea+b(lnb1)e^a-b \geq b(a-\ln b) \Rightarrow ab \leq e^a+ b(\ln b -1).  \space Case 2 If lnbx0>a\ln b \geq x \geq 0 >a then in this interval bex b\geq e^x. Thus alnbbdxalnbexdx\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(lnba)beaabea+b(lnb1)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 - 5 years, 7 months ago

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thanks! :)

hemang sarkar - 5 years, 3 months ago

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