9
$\begingroup$

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...

$\endgroup$
  • 5
    $\begingroup$ If you look at the binomial formula, you will see it pretty shortly. $\endgroup$ – BBischof Aug 2 '10 at 0:59
  • 5
    $\begingroup$ You don't even need the binomial. It is pretty clear that these terms have to occur in the expansion. $\endgroup$ – Casebash Aug 2 '10 at 1:05
14
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Please continue. I expanded the left side, and it has an a^n summed with a b^n term (along with a whole bunch of other terms). Is that proof enough? $\endgroup$ – Matt Aug 2 '10 at 1:23
  • $\begingroup$ Matt: remember, a and b are both positive (that should answer your question) $\endgroup$ – Jamie Banks Aug 2 '10 at 1:33
13
$\begingroup$

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 $a^n + b^n$.

$\endgroup$
10
$\begingroup$

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 $b$?

$\endgroup$
10
$\begingroup$

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 $n \geq 1$.

$\endgroup$
  • 3
    $\begingroup$ So simple!${}{}$ $\endgroup$ – TonyK Nov 13 '11 at 11:52
5
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ A precalculus answer? But yours is the only answer that uses calculus, while the others are more elementary! :-) (BTW, it should be $n((a+b)^{n-1}-a^{n-1})$.) $\endgroup$ – ShreevatsaR Aug 2 '10 at 7:12
  • 1
    $\begingroup$ Sorry I am French and we make no distinction between "analyse" (calculus) and anything like "pré-analyse" (precalculus). But now thanks to you, I know the difference. $\endgroup$ – marwalix Aug 2 '10 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.