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
5 years ago
Easy Math Editor
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
Sort by:
Top NewestThis is actually a really interesting topic that I have never heard of! Thanks for sharing!
Log in to reply
Very nice problems keep it up nicolae
Log in to reply
Sir, Can You give Solution to Problem 3,,I am not able to do it. I have done With 2nd one
Log in to reply
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.
Log in to reply
thanks! :)
Log in to reply