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

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

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 $n \geq 1$.

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

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 7

Induction.

For n=1 it is trivially true

Assume true for n=k i.e. (a + b)^k >= a^k + b^k

Consider case n=k+1 (a+b)^(k+1)=(a+b)(a+b)^k>=(a+b)(a^k+b^k)=a^(k+1)+b^(k+1)+ab^k+ba^k>=a^(k+1)+b^(k+1) as required