Let a > b > 0 and n ∈ N. Prove that(an + cn)1/n − (bn + cn)1/n ≤ a − b for all c ≥ 0

To prove the inequality $(a^n + c^n)^{1/n} - (b^n + c^n)^{1/n} \leq a - b$ for all $c \geq 0$, given that $a > b > 0$ and $n \in \mathbb{N}$, we can follow the steps outlined below.

1. Break Down the Problem

• We need to show that the difference between the two expressions, $(a^n + c^n)^{1/n}$ and $(b^n + c^n)^{1/n}$, is less than or equal to $a - b$.
• Consider $f(x) = (x^n + c^n)^{1/n}$ for $x \geq 0$.

2. Relevant Concepts

• We will use the Mean Value Theorem (MVT), which states that if $f(x)$ is continuous on $[b, a]$ and differentiable on $(b, a)$, then there exists a $c \in (b, a)$ such that: $f(a) - f(b) = f'(c)(a - b)$
• The derivative $f'(x)$ can be computed as follows: $f'(x) = \frac{d}{dx} (x^n + c^n)^{1/n} = \frac{n}{n}(x^n + c^n)^{(1/n)-1} \cdot nx^{n-1}$

3. Analysis and Detail

• Let's compute $f'(x)$: $f'(x) = \frac{x^{n-1}}{(x^n + c^n)^{(n-1)/n}}$
• By the MVT, we have: $(a^n + c^n)^{1/n} - (b^n + c^n)^{1/n} = f'(d)(a - b)$ for some $d \in (b, a)$.
• Since $f'(x)$ is a positive function for $x > 0$, and since $a > b > 0$, we can conclude: $f'(d) \geq \frac{b^{n-1}}{(b^n + c^n)^{(n-1)/n}} > 0$
• Therefore: $(a^n + c^n)^{1/n} - (b^n + c^n)^{1/n} \leq f'(d)(a - b) < a - b$

Thus, we get: $(a^n + c^n)^{1/n} - (b^n + c^n)^{1/n} \leq a - b$

4. Verify and Summarize

• We have derived the result using the Mean Value Theorem and assessed the behavior of $f'(x)$.
• This shows that the difference is indeed less than or equal to $a - b$ for any $c \geq 0$ and given $a > b > 0$.

Final Answer

Therefore, it is proved that: $(a^n + c^n)^{1/n} - (b^n + c^n)^{1/n} \leq a - b \quad \forall c \geq 0$