Animated Solution for Mathematics - Differentiation: Show that $1+x\ln (x+\sqrt{x^2+1})\ge \sqrt{1+x^2}$ for all $x\ge 0$.

Welcome, future engineer. Today, we are not just solving an inequality; we are embarking on a journey of mathematical discovery. When you look at an expression like $1+x\ln (x+\sqrt{x^2+1})\ge \sqrt{1+x^2}$, it might seem daunting.

But in the world of JEE Advanced, we don't fear complexity; we embrace it. We look for the underlying structure. The first step is to define a function:

By moving everything to one side, we are essentially asking: "How does the gap between these two expressions change as $x$ grows?"

Before we dive into the deep waters of calculus, we must find our anchor. We evaluate $f(0)$.

Substituting $x=0$ into our function, the middle term vanishes because it is multiplied by zero, and the last term becomes $\sqrt{1}=1$. Thus:

This is our starting point. Our function begins exactly at the origin. This is a crucial piece of information—it tells us that if we can prove the function is always increasing, it can never dip below zero.

Now, let us find the rate of change. We differentiate $f(x)$ with respect to $x$. The derivative of the constant $1$ is $0$.

For the second term, $x\ln (x+\sqrt{x^2+1})$, we must use the product rule. The derivative of $x$ is $1$, so we have $1\cdot \ln (x+\sqrt{x^2+1})$.

Then, we add $x$ multiplied by the derivative of the natural log. The derivative of $\ln (u)$ is $\frac{1}{u}\cdot u^{\prime }$. When you differentiate $x+\sqrt{x^2+1}$, you get:

When you multiply this by $\frac{1}{x+\sqrt{x^2+1}}$, the terms cancel out beautifully, leaving us with $\frac{1}{\sqrt{x^2+1}}$.

Finally, we differentiate the last term, $\sqrt{1+x^2}$. Using the chain rule, this becomes $\frac{x}{\sqrt{1+x^2}}$.

Now, look at our complete derivative $f^{\prime }(x)$: we have the log term, plus $x\cdot \frac{1}{\sqrt{x^2+1}}$, minus $\frac{x}{\sqrt{1+x^2}}$. The last two terms are identical! They cancel each other out completely.

We are left with:

For any $x>0$, we know that $\sqrt{x^2+1}>1$, which means $x+\sqrt{x^2+1}>1$. The natural log of any number greater than $1$ is positive.

Therefore, $f^{\prime }(x)>0$ for all $x>0$. Since our function starts at $0$ and its slope is always positive, it must be strictly increasing.

It never looks back and it never dips below zero. We have successfully proven that $f(x)\ge 0$ for all $x\ge 0$. This is the elegance of calculus—turning a complex inequality into a simple story of motion.