Animated Solution for Mathematics - Sequence and Series: If $a,b,c$ are positive real numbers. Then prove that $(a+1{)}^7(b+1{)}^7(c+1{)}^7>7^7{a}^4{b}^4{c}^4$

A Journey into AM-GM

Welcome, fellow explorers of mathematics. Today, we are not just solving an inequality; we are uncovering the hidden architecture of numbers.

We are tasked with proving that for any positive real numbers $a,b,c$, the expression $(a+1{)}^7(b+1{)}^7(c+1{)}^7$ is strictly greater than $7^7{a}^4{b}^4{c}^4$.

At first glance, this looks like a daunting wall of exponents. But remember, in the world of JEE Advanced, complexity is often just a mask for a beautiful, underlying simplicity. Our key to unlocking this is the Arithmetic Mean-Geometric Mean ($AM\ge GM$) inequality.

The core of this problem lies in how we view the term $(a+1)$. If we apply $AM\ge GM$ directly to $a$ and $1$, we get $\frac{a+1}{2}\ge \sqrt{a}$, which gives $(a+1{)}^2\ge 4a$.

This is true, but it doesn't help us reach the power of $7$ or the $a^4$ term. We need to be more surgical. We need $a^4$ in our product, which means we need $4$ terms of $a$.

Since we are dealing with a power of $7$, we need $7$ terms in total. This forces us to introduce $3$ constant terms. We partition $(a+1)$ as follows:

This is the 'Spark' of the problem. By splitting $a$ into four parts and $1$ into three, we create a structure that, when multiplied, perfectly yields $a^4$ and a constant.

Now, let us apply the $AM\ge GM$ inequality to these $7$ terms:

This simplifies to:

Raising both sides to the power of $7$, we get:

Let $K=4^4\cdot 3^3=256\cdot 27=6912$. Thus, $(a+1{)}^7\ge \frac{7^7}{6912}{a}^4$.

Because of the symmetry of the problem, we can immediately write the same for $b$ and $c$:

Now, we multiply these three inequalities together. The left side becomes $(a+1{)}^7(b+1{)}^7(c+1{)}^7$, and the right side becomes:

To complete the proof, we must show that $\frac{7^{21}}{(6912{)}^3}>7^7$. Dividing by $7^7$, we need to verify that:

Calculating these values, $7^{14}=49^7\approx 6.78\times 10^{11}$, while $6912^3\approx 3.30\times 10^{11}$.

Since the numerator is clearly larger than the denominator, the inequality holds. We have successfully navigated the complexity and arrived at the truth. Keep practicing this art of partitioning; it is the secret weapon of every top-tier mathematician.