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 $ab\leq e^a+b(\ln b -1).$