Prove: $(a + b)^{n} \geq a^{n} + b^{n}$

Struggling with yet another proof:

Prove that, for any positive integer $n: (a + b)^n \geq a^n + b^n$ for all $a, b > 0:$

I wasted $3$ pages of notebook paper on this problem, and I'm getting nowhere slowly. So I need some hints.

$1.$ What technique would you use to prove this (e.g. induction, direct, counter example)

$2.$ Are there any tricks to the proof? I've seen some crazy stuff pulled out of nowhere when it comes to proofs...

Solution 1:

Hint: Use the binomial theorem.

This states that $(a + b)^n = \sum \limits_{k = 0}^n {n \choose k} a^{n-k} b^k = a^n + b^n + \sum \limits_{k=1}^{n-1} {n \choose k} a^{n-k} b^k$.

Now, note that every term in the second sum is positive; this is because a, b, and the binomial coefficients are all positive. Therefore, $(a+b)^n=a^n+b^n+\text{ (sum of positive terms) }\geqslant a^n+b^n\;.$

Solution 2:

This follows directly from the binomial theorem. Alternatively, you can prove it inductively (which is probably more fun): suppose the inequality true for $n-1$. Then $(a+b)^n = (a+b)(a+b)^{n-1} \geq (a+b)(a^{n-1} + b^{n-1})$ by the inductive hypothesis. So $(a+b)^n \geq a(a^{n-1}+ b^{n-1}) + b(b^{n-1} + a^{n-1})$, and this is at least $a^n + b^n$.