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

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 = aⁿ + bⁿ + (sum of positive terms) >= aⁿ + bⁿ.

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 $a^n + b^n$.

It might also be helpful for you to think a little about the geometry of the inequality.

For $n=2$, find a way to put an $a \times a$ square and a $b \times b$ square into a $(a+b) \times (a+b)$ square without any overlaps. For $n=3$, see if you can fit an $a \times a \times a$ cube and a $b \times b \times b$ cube within an $(a+b) \times (a+b) \times (a+b)$ cube without overlaps.

Next, the notion of having more than three dimensions might seem a little weird, but think of the box in $n$ dimensions whose sides have length $a+b$. Can you fit two boxes within it, one with side length $a$ and one with side length $b$?