Animated Solution for Mathematics - Differentiation: Prove that for $x\in [0,\pi /2],\sin x+2x\ge \frac{3x(x+1)}{\pi }$. Explain the identity if any used in the proof.

A Journey into Concavity

Welcome, fellow traveler of the mathematical landscape. Today, we are going to tackle a problem that might look like a simple inequality, but it is actually a gateway into the elegant world of calculus-based proofs.

We are asked to prove that for $x\in [0,\frac{\pi }{2}]$, the inequality $\sin x+2x\ge \frac{3x(x+1)}{\pi }$ holds true. This is not just about plugging in numbers; it is about understanding the soul of the function.

The Setup

When you see an inequality in a JEE Advanced paper, your first instinct should be to simplify the battlefield. We want to prove that one expression is greater than or equal to another.

Let us define a new function, $f(x)$, that captures the difference between these two sides:

Our goal is now crystal clear: we need to show that $f(x)\ge 0$ for all $x$ in the interval $[0,\frac{\pi }{2}]$. By moving everything to one side, we have transformed a comparison problem into a root-finding or positivity problem.

It is much easier to analyze the behavior of a single function than to juggle two separate expressions.

The Boundary Check

Before we dive into the deep waters of derivatives, let us test the boundaries. What happens at the start of our interval, $x=0$?

Excellent! The function starts right at the origin. Now, let us check the other end, $x=\frac{\pi }{2}$:

Simplifying this, we get:

Since $\pi \approx 3.14$, $\frac{\pi }{4}$ is roughly $0.785$. Subtracting $0.5$ gives us a positive value. So, our function starts at zero and ends at a positive value.

The Derivative Analysis

To understand the shape of our curve, we must look at its rate of change. Let us find the first derivative, $f^{\prime }(x)$:

This tells us how the function is increasing or decreasing. But to truly understand the 'bend' of the curve, we need the second derivative, $f^{\prime \prime }(x)$:

Look closely at this expression. In the interval $[0,\frac{\pi }{2}]$, $\sin x$ is always non-negative. Therefore, $-\sin x$ is non-positive.

Since $\frac{6}{\pi }$ is a positive constant, $f^{\prime \prime }(x)$ is strictly negative throughout the entire interval. It means our function is concave down.

The Geometric Insight

This is the moment of truth. A function that is concave down on an interval always lies above the secant line connecting its endpoints. Think of it like an inverted bowl; the curve bows upwards, staying above the straight line that connects the start and end points.

Since our secant line connects $(0,0)$ and $(\frac{\pi }{2},f(\frac{\pi }{2}))$, and we know $f(\frac{\pi }{2})>0$, the secant line itself is non-negative for all $x\in [0,\frac{\pi }{2}]$.

Because our curve is concave down and lies above this non-negative secant line, $f(x)$ must also be non-negative. We have proven that $\sin x+2x\ge \frac{3x(x+1)}{\pi }$ by leveraging the power of concavity.