Animated Solution for Mathematics - Differentiation: Show that $2\sin x+\tan x\ge 3x$ where $0\le x<\pi /2$.

Mastering Trigonometric Inequalities

Have you ever looked at an inequality like $2\sin x+\tan x\ge 3x$ and felt a bit intimidated? It looks like a jumble of trigonometric functions fighting against a linear term.

But in the world of JEE Advanced, these problems are not just algebraic exercises; they are stories about how functions grow. Today, we are going to peel back the layers of this inequality and see the beautiful logic hidden underneath.

Defining Our Landscape

The first step in any battle is to define the terrain. We want to prove that $2\sin x+\tan x\ge 3x$ for $x\in [0,\pi /2)$.

Instead of staring at both sides, let's bring everything to one side. We define a new function:

Our mission is simple: if we can prove that $f(x)\ge 0$ for the entire interval, we have won the war.

Starting at Zero

Every journey begins with a single step. Let's check our function at the starting point, $x=0$.

Substituting this into our function, we get:

Since $\sin (0)=0$ and $\tan (0)=0$, we find that $f(0)=0$. This is our anchor. Our function starts exactly at the origin.

If we can prove that this function is always climbing—that it never turns back down—then it must always be greater than or equal to zero.

The Derivative

To understand if our function is always climbing, we need to look at its 'velocity,' which in calculus terms is the derivative, $f^{\prime }(x)$. Let's differentiate $f(x)$ with respect to $x$:

Applying the standard derivatives, we get:

Now, the question becomes: is this derivative always non-negative? If $f^{\prime }(x)\ge 0$, then our function $f(x)$ is monotonically increasing.

A Moment of Elegance

We need to show that $2\cos x+{\sec }^{2}x-3\ge 0$, which is equivalent to showing $2\cos x+{\sec }^{2}x\ge 3$. This is where the Arithmetic Mean-Geometric Mean (AM-GM) inequality shines.

The AM-GM inequality states that for positive numbers, the arithmetic mean is always greater than or equal to the geometric mean. We have three terms: $\cos x$, $\cos x$, and ${\sec }^{2}x$. Let's apply AM-GM:

Look at the product inside the cube root:

The magic happens here! The terms cancel out perfectly, leaving us with:

Multiplying by 3, we get $2\cos x+{\sec }^{2}x\ge 3$.

We have done it. We have proven that $f^{\prime }(x)=(2\cos x+{\sec }^{2}x)-3\ge 3-3=0$.

Because the derivative is always non-negative, the function $f(x)$ is monotonically increasing on the interval $[0,\pi /2)$. Since it starts at $f(0)=0$ and only ever increases, it must be true that $f(x)\ge 0$ for all $x$ in the interval.

Thus, $2\sin x+\tan x-3x\ge 0$, which leads us directly to our goal: $2\sin x+\tan x\ge 3x$.

Take a moment to appreciate this. We didn't just solve an inequality; we mapped the behavior of a function and used the elegance of AM-GM to prove its path. Keep practicing this mindset—look for the function, find the anchor, and analyze the growth. You've got this!