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(2sinx)>21x(π−x)−725π2,∀x∈(6π,2π)
Solution. We know the very basic inequality tanx>x for 0<x<2π. Now if we substitute x with 2π−x we'll obtain cotx>2π−x. Now integrating our inequality we obtain
∫6πxcottdt>∫6πx(2π−t)dt.
Calculating we get
ln(sinx)−ln21>(2πx−2x2)−(2π⋅6π−21(6π2))⇔
⇔ln(2sinx)>21x(π−x)−725π2.
Problem 2. For 0<x<2π prove the following inequality
sinx≤2x(π−x).
Problem 3. For b≥1 prove the following inequality
ab≤ea+b(lnb−1).
Easy Math Editor
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Top NewestThis is actually a really interesting topic that I have never heard of! Thanks for sharing!
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Very nice problems keep it up nicolae
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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
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I could only solve it by considering two cases. Case 1. If a>x>lnb≥0 then in this interval ex≥b≥1. Thus ∫lnbaexdx≥∫lnbabdx. On expanding we have ea−b≥b(a−lnb)⇒ab≤ea+b(lnb−1). Case 2 If lnb≥x≥0>a then in this interval b≥ex. Thus ∫alnbbdx≥∫alnbexdx. Which on expanding gives us b(lnb−a)≥b−ea⇒ab≥ea+b(lnb−1). PS: Let me know if you find a mistake or an easier way, all criticisms are welcomed.
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thanks! :)
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