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

Question

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...

Answer 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, $(a+b)^n=a^n+b^n+\text{ (sum of positive terms) }\geqslant a^n+b^n\;.$

Answer 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 $a^n + b^n$.

Answer 3

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 $b$?

Answer 4

Let's have a precalculus answer: Consider the function of $a$ depending on the parameter $b$ that is $f_b(a)=(a+b)^n-a^n-b^n$. Its derivative relative to $a$ is $f'_b(a)=n((a+b)^{n-1}-a^{n-1})$ because $b>0$ $f'_b(a,b)$ is nonnegative and $f_b$ is an increasing function of $a$ and you can conclude.

Answer 5

Here is another way to look at the inequality. Pick the larger of $a^n$ and $b^n$ and divide through by that quantity. This reduces the problem to showing that $(1+r)^n \ge 1 + r^n$ for some positive real number $r \le 1$. If $r < 1$, we have $r^n < r$, so $1 + r^n < 1 + r < (1+r)^n$. I leave the case $r = 1$ (and $n$ rational and $\ge 1$) to others.

Answer 6

Induction.

For $n=1$ it is trivially true

Assume true for $n=k$

i.e. $$(a + b)^k \ge a^k + b^k$$

Consider case $n=k+1$ \begin{align}&(a+b)^{k+1} =(a+b)(a+b)^k\\ &\ge(a+b)(a^k+b^k)\\ &=a^{k+1}+b^{k+1}+ab^k+ba^k\\ &\ge a^{k+1}+b^{k+1}\end{align}

Answer 7

You can write $n=m+1$ where $m \geq 0$, then

$(a+b)^n = (a+b)^{m+1} = (a+b) (a+b)^m = a(a+b)^m +b(a+b)^m \geq a^{m+1} + b^{m+1}$

no induction and works for any real $n \geq 1$.