Animated Solution for Mathematics - Differentiation: The real number $x$ when added to its inverse gives the minimum value of the sum at $x$ equal to

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to explore a problem that seems deceptively simple but hides a beautiful geometric truth.

We are looking for the minimum value of the sum of a real number and its inverse. Let us define our function as:

Imagine you are standing on a graph of this function. When $x$ is a very large positive number, the term $x$ dominates, and the function grows towards infinity.

When $x$ is a very small positive number (close to zero), the term $\frac{1}{x}$ explodes towards infinity. There is a constant tug-of-war between these two terms.

The minimum value occurs at the exact point where these two forces are in perfect balance. This is the heart of the problem.

If we plot $f(x)=x+\frac{1}{x}$ for $x>0$, we see a curve that descends from infinity, hits a local minimum, and then rises again. This is the 'dip' we are searching for.

If we look at the negative branch ($x<0$), the function behaves differently. It goes down towards negative infinity as $x$ becomes more negative.

Because the function does not have a lower bound on the negative side, the question of a 'minimum' only makes sense when we restrict our focus to the positive branch, where $x>0$.

To find this minimum without the heavy machinery of calculus, we turn to the Arithmetic Mean-Geometric Mean (AM-GM) inequality. This theorem is a cornerstone of algebraic optimization.

It states that for any two positive real numbers $a$ and $b$, the following holds:

Let us apply this to our function. We set $a=x$ and $b=\frac{1}{x}$. Substituting these into the inequality, we get:

The magic happens right here. The product $x\cdot \frac{1}{x}$ simplifies perfectly to $1$.

Thus, the inequality becomes $\frac{x+\frac{1}{x}}{2}\ge \sqrt{1}$, which simplifies to:

This tells us that the sum can never be less than $2$.

The question asks for the value of $x$ where this minimum occurs. The AM-GM inequality tells us that the minimum value is achieved if and only if the two terms are equal: $a=b$.

So, we set $x=\frac{1}{x}$. Multiplying both sides by $x$, we get:

Since we are working on the positive branch, we take the positive root: $x=1$.

At $x=1$, the sum is $1+\frac{1}{1}=2$. We have found our balance point.

This is the beauty of mathematics—a complex-looking problem reduced to a simple, elegant point of symmetry. Keep this tool in your arsenal; the AM-GM inequality will serve you well in many more JEE challenges to come. The minimum value is 2.