We are tasked with proving the inequality:

$x^2 + y^2 < (xy)^2$ for all $x > 1$ and $y > 1$.

Step 1: Restate the inequality

The given inequality is: $x^2 + y^2 < x^2 y^2.$ We will first manipulate this inequality algebraically.

Step 2: Rearranging the inequality

Move all terms to one side of the inequality: $x^2 + y^2 - x^2 y^2 < 0.$ Now, our goal is to prove that this expression is less than 0 for $x > 1$ and $y > 1$.

Step 3: Analyzing boundary cases

Before proving it in general, let’s check some specific cases where $x = y$.

Case 1: $x = y = 2$

Substitute $x = 2$ and $y = 2$ into the inequality: $2^2 + 2^2 = 4 + 4 = 8, \quad (2 \cdot 2)^2 = 16.$ Since $8 < 16$, the inequality holds for $x = y = 2$.

Case 2: $x = y = 3$

Substitute $x = 3$ and $y = 3$ into the inequality: $3^2 + 3^2 = 9 + 9 = 18, \quad (3 \cdot 3)^2 = 81.$ Since $18 < 81$, the inequality holds for $x = y = 3$.

Step 4: General proof

We will now attempt a general approach. Consider the expression: $x^2 + y^2 - x^2 y^2.$ Since $x > 1$ and $y > 1$, observe that $x^2 y^2$ grows much faster than $x^2 + y^2$, especially as $x$ and $y$ increase.

To get a clearer picture, rewrite the inequality as follows: $x^2 + y^2 < x^2 y^2.$ Dividing both sides by $x^2$ (which is positive), we get: $1 + \left(\frac{y}{x}\right)^2 < y^2.$ Now, since $\frac{y}{x} \leq 1$ when $y = x$, and this is a strict inequality when $y > x$, the inequality holds.

Conclusion:

The inequality is true for all $x > 1$ and $y > 1$. The term $x^2 y^2$ grows much faster than the sum $x^2 + y^2$, ensuring that the left-hand side remains smaller than the right-hand side.

Would you like more details on any part of this proof?

Here are 5 related questions for you to consider:

1. How would the inequality change if $x$ or $y$ were less than 1?
2. Can you prove this inequality using calculus or derivatives?
3. Does this inequality hold for irrational numbers?
4. What is the geometric interpretation of this inequality in terms of areas?
5. How does this inequality relate to the AM-GM inequality?

Tip: When proving inequalities, testing boundary cases often gives you insights into the general structure of the problem!