Problem 1697

To prove that the two statements are equivalent, we need to show that each statement implies the other.

1. Assume (1): $\sqrt{a} + 1 > \sqrt{b}$

We need to show that this implies (2): $\forall x > 1, \ ax + \frac{x}{x - 1} > b$.

Let's start by manipulating the inequality $\sqrt{a} + 1 > \sqrt{b}$.

$$\sqrt{a} + 1 > \sqrt{b}$$

Subtract 1 from both sides:

$$\sqrt{a} > \sqrt{b} - 1$$

Square both sides to eliminate the square root:

$$a > (\sqrt{b} - 1)^2$$

Expand the right-hand side:

$$a > b - 2\sqrt{b} + 1$$

Now, consider the expression $ax + \frac{x}{x - 1}$ for $x > 1$:

$$ax + \frac{x}{x - 1}$$

We need to show that this is greater than $b$. Let's rewrite $\frac{x}{x - 1}$:

$$\frac{x}{x - 1} = 1 + \frac{1}{x - 1}$$

So,

$$ax + \frac{x}{x - 1} = ax + 1 + \frac{1}{x - 1}$$

Since $x > 1$, $\frac{1}{x - 1}$ is positive. Therefore,

$$ax + 1 + \frac{1}{x - 1} > ax + 1$$

We need to show that $ax + 1 > b$. Using the inequality $a > b - 2\sqrt{b} + 1$:

$$ax + 1 > (b - 2\sqrt{b} + 1)x + 1$$

Simplify the right-hand side:

$$ax + 1 > bx - 2x\sqrt{b} + x + 1$$

Subtract $bx$ and $1$ from both sides:

$$ax - bx > -2x\sqrt{b} + x$$

Factor out $x$:

$$(a - b)x > -2x\sqrt{b} + x$$

Since $x > 1$, we can divide both sides by $x$:

$$a - b > -2\sqrt{b} + 1$$

This is true because $a > b - 2\sqrt{b} + 1$. Hence, (1) implies (2).

2. Assume (2): $\forall x > 1, \ ax + \frac{x}{x - 1} > b$

We need to show that this implies (1): $\sqrt{a} + 1 > \sqrt{b}$.

Consider $x \to 1^+$. As $x$ approaches 1 from the right, $\frac{x}{x - 1}$ becomes very large. Therefore, for $x$ very close to 1, the term $\frac{x}{x - 1}$ dominates.

Given $ax + \frac{x}{x - 1} > b$, let $x = 1 + \epsilon$ where $\epsilon$ is a small positive number:

$$a(1 + \epsilon) + \frac{1 + \epsilon}{\epsilon} > b$$

Simplify the fraction:

$$a + a\epsilon + \frac{1}{\epsilon} + 1 > b$$

As $\epsilon \to 0$, $\frac{1}{\epsilon} \to \infty$. Therefore, the inequality holds for very small $\epsilon$:

$$a + \frac{1}{\epsilon} + 1 > b$$

This implies that $a + 1 > b$ for the inequality to hold as $\epsilon \to 0$. Now, consider the original inequality $\sqrt{a} + 1 > \sqrt{b}$:

$$\sqrt{a} + 1 > \sqrt{b}$$

Square both sides:

$$(\sqrt{a} + 1)^2 > b$$

Expand the left-hand side:

$$a + 2\sqrt{a} + 1 > b$$

Since $a + 1 > b$, we have:

$$2\sqrt{a} > 0$$

This is always true for positive $a$. Hence, (2) implies (1).

Therefore, the two statements are equivalent.

$\blacksquare$