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I'm currently solving problems involving AM-GM and I am stuck on this problem.

Prove that if $a,b,c>0$ and $n,k \in \mathbb{Z}^{+}, n>k$ that $a^n + b^n \geq a^{n-k}b^{k} + a^{k}b^{n-k}$

I've attempted the case when $n=3$ and found that $a^3 + b^3 \geq ab^2+a^2b$. However, the problem above is even more general than that.

chuckong083608
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2 Answers2

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Your exponents are already very suggestive - $a^kb^{n-k}$ is asking you to procure $k$ and $n-k$, $a^n$ and $b^n$'s, respectively. The most natural way comes to mind is then: $$ka^n + (n-k)b^n = \underbrace{(a^n+\dots + a^n)}_{k} + \underbrace{(b^n+\dots+b^n)}_{n-k}\geq n\sqrt[n]{a^{kn}b^{n(n-k)}} = na^kb^{n-k}.$$

Can you fill in the details now ? Alternatively, if you had done the case $n=3, k=2$, then you have essentially solved the general case except you just had to write it in the general form like I did.

dezdichado
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    Yes. Similarly, $(n-k)a^n + kb^n = \underbrace{(a^n+\dots + a^n)}{n-k} + \underbrace{(b^n+\dots+b^n)}{k}\geq n\sqrt[n]{a^{n(n-k)}b^{kn}} = na^{n-k}b^{k}.$

    Finally, add both inequalities and divide both sides by n.

    Thank you so much!

     – chuckong083608 Jun 12 '23 at 14:57
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An alternative to using AM-GM would be a simple factorization, WLOG, assume that $a\ge b$, then $$a^n-a^{k}b^{n-k}+ b^n - a^{n-k}b^k = (a^k-b^k)(a^{n-k}-b^{n-k})\ge 0.$$

ChengYiin
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