Natural numbers inequality $na^{n-1}b\leq(n-1)a^n+b^n$ by induction

Question

Let $a$ and $b$ be arbitrary natural numbers and $n$ some positive integer. How to prove the inequality $$na^{n-1}b\leq(n-1)a^n+b^n$$ by induction for all $n$? This is related to this result, and, using the method described there, I can prove the inequality when $a\leq b$, but I'm supposed to prove for arbitrary natural numbers. What to do in the case $b<a$?

Answer 1

It's just AM-GM: $$(n-1)a^n+b^n\geq n\sqrt[n]{(a^{n})^{n-1}b^n}=na^{n-1}b.$$

Also, by the assumption of an induction and by Rearrangement we obtain: $$na^{n+1}+b^{n+1}=(n-1)a^{n+1}+ab^n+a^{n+1}+b^{n+1}-ab^n\geq$$ $$\geq na^{n-1}ba+a^{n+1}+b^{n+1}-ab^n=$$ $$=(n+1)a^nb+a^{n+1}+b^{n+1}-a^nb-ab^n\geq(n+1)a^nb.$$ We can get the last inequality also by the following way: $$a^{n+1}+b^{n+1}-a^nb-ab^n=(a^n-b^n)(a-b)\geq0.$$

Answer 2

Assume we know that $ka^{k-1}b\le(k-1)a^k+b^k$.

$ka^kb \le (k-1)a^{k+1} + ab^k$

$(k+1)a^kb \le (k-1)a^{k+1} + [ab^k + a^kb]$

$=(k-1)a^{k+1} + [ab^k + a^ka + a^kb-a^ka]=ka^{k+1}+[ab^k + a^k(b-a)]$

$=ka^{k+1} + b^{k+1} + (ab^k- bb^k)+a^k(b-a)$

$=(ka^{k+1} + b^{k+1}) + [b^k(a-b) + a^k(b-a)]= (ka^{k+1} + b^{k+1}) + (a-b)(b^k-a^k)$.

Now $a-b \ge 0 \iff b^k - a^k \le 0$ so $(a-b)(b^k-a^k) \le 0$ so

$(k+1)a^kb \le (ka^{k+1} + b^{k+1}) + (a-b)(b^k-a^k) \le ka^{k+1} + b^{k+1}$.