Proving $\frac{a^n}{b^m} \ge \frac{n a^{n-m}-m b^{n-m}}{n-m}$ and $a^n+b^n \ge a^{n-k}b^{k}+a^k b^{n-k}$

Question

I want to prove the following two inequalities:

1) If $a, b >0$ and $n, m \in \mathbb{N}$, $n>m$, then $$\frac{a^n}{b^m} \ge \frac{n a^{n-m}-m b^{n-m}}{n-m}$$

2) If $a, b > 0$, $n, k \in \mathbb{N}$, $1\le k\le n$, then $$a^n+b^n \ge a^{n-k}b^{k}+a^k b^{n-k}$$

I think that they should be proveable by straightforward calculations since I found then in the introductory section of a textbook, but I cannnot manage to prove them. I got that $1) \iff na^n(a^m-b^m)\ge m a^m (a^n-b^n)$, but this is not obviously true, some more work needs to be done. I tried something similar for $2)$, but it also didn't work out.

Answer 1

The second: $$a^n+b^n-a^{n-k}b^k-a^kb^{n-k}=a^{n-k}(a^k-b^k)-b^{n-k}(a^k-b^k)=(a^k-b^k)(a^{n-k}-b^{n-k})\geq0.$$ The first gives:

Let $\frac{a}{b}=x.$

Thus, $x>0$ and by your work we need to prove that: $$\frac{x^m-1}{m}\geq\frac{x^n-1}{n},$$ which seems reversed.

Indeed, let $f(x)=\frac{x^n-1}{n}-\frac{x^m-1}{m}.$

Thus, $$f'(x)=x^{n-1}-x^{m-1}=x^{m-1}(x^{n-m}-1),$$ which gives $x_{min}=1$ and $$f(x)\geq f(1)=0$$ and the inequality is indeed reversed.