IIT-JEE 2004Question 1
Prove that for $x\in\left[0,\dfrac{\pi}{2}\right]$, $\sin x+2x\ge\dfrac{3x(x+1)}{\pi}$. Explain the identity, if any, used in the proof.
Differentiation and its Application
Answerproved
Solution
Concept. Use Jordan's inequality $\sin x\ge\frac{2x}{\pi}$ on $\left[0,\frac\pi2\right]$, then a simple bound on $x$. Formulas used. (1) Jordan's inequality: $\sin x\ge\dfrac{2x}{\pi}$ for $x\in\left[0,\frac\pi2\right]$Step 1 - Prove (1). Let $u(x)=\sin x-\frac{2x}{\pi}$. Then $u(0)=u\left(\frac\pi2\right)=0$ and $u''=-\sin x\le0$, so $u$ is concave: its graph lies above the chord joining its end points, which is the $x$-axis. So $u\ge0$. (This is the inequality, "identity", used.) Step 2 - Use (1). $$\sin x+2x-\frac{3x(x+1)}{\pi}\ \ge\ \frac{2x}{\pi}+2x-\frac{3x^2+3x}{\pi}=x\left(2-\frac1\pi-\frac{3x}{\pi}\right).$$ Step 3 - Bound $x$. $x\le\frac\pi2$ gives $\frac{3x}{\pi}\le\frac32$, so the bracket is at least $\frac12-\frac1\pi>0$. With $x\ge0$ the whole expression is $\ge0$. Hence proved.