Inequalities for Real Numbers and Positive Integer n

The problem asks us to identify the true statements among the given options for real numbers x and y such that $0 < y \le x$, and for a positive integer n. The statements involve comparing the term $x^n - y^n$ with expressions involving $n$, $x$, $y$, and $x-y$.

Deriving Inequalities using Calculus

We can use calculus, specifically the Mean Value Theorem, to derive these inequalities. Consider the function $f(t) = t^n$ for $t$ in the interval $[y, x]$. Since n is a positive integer, $f(t)$ is continuous on $[y, x]$ and differentiable on $(y, x)$.

By the Mean Value Theorem, there exists some value $c$ in the interval $(y, x)$ such that: $$ \frac{f(x) - f(y)}{x - y} = f'(c) $$ The derivative of $f(t) = t^n$ is $f'(t) = nt^{n-1}$. Substituting this into the Mean Value Theorem equation: $$ \frac{x^n - y^n}{x - y} = nc^{n-1} $$ Multiplying both sides by $(x - y)$ (which is non-negative since $x \ge y$), we get: $$ x^n - y^n = nc^{n-1} (x - y) $$

Now, we use the fact that $y < c < x$. Since $n$ is a positive integer, $n-1$ is a non-negative integer. The function $g(t) = t^{n-1}$ is non-decreasing for $t > 0$. Given $y > 0$, we have:

• Since $y < c$, it follows that $y^{n-1} \le c^{n-1}$.
• Since $c < x$, it follows that $c^{n-1} \le x^{n-1}$.

Multiplying these inequalities by the non-negative term $n(x - y)$, we get: $$ ny^{n-1}(x - y) \le nc^{n-1}(x - y) \le nx^{n-1}(x - y) $$ We already established that $x^n - y^n = nc^{n-1}(x - y)$. Substituting this into the compound inequality: $$ ny^{n-1}(x - y) \le x^n - y^n \le nx^{n-1}(x - y) $$

First Inequality: $\text{ny}^\text{n-1} \text{(x - y)} \le \text{x}^\text{n} \text{- y}^\text{n}$

From the derivation above, the left side of the inequality chain is $ny^{n-1}(x - y) \le x^n - y^n$. This statement is true under the given conditions.

Second Inequality: $\text{nx}^\text{n-1} \text{(x - y)} \ge \text{x}^\text{n} \text{- y}^\text{n}$

From the derivation above, the right side of the inequality chain is $x^n - y^n \le nx^{n-1}(x - y)$, which can be rewritten as $nx^{n-1}(x - y) \ge x^n - y^n$. This statement is also true under the given conditions.

Conclusion

Based on the Mean Value Theorem applied to the function $f(t) = t^n$ on the interval $[y, x]$, we derived the compound inequality $ny^{n-1}(x - y) \le x^n - y^n \le nx^{n-1}(x - y)$. This shows that both the statements:

• $\text{ny}^\text{n-1} \text{(x - y)} \le \text{x}^\text{n} \text{- y}^\text{n}$
• $\text{nx}^\text{n-1} \text{(x - y)} \ge \text{x}^\text{n} \text{- y}^\text{n}$

are true for real numbers x, y with $0 < y \le x$ and any positive integer n.