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

Having problem with yet an additional evidence:

Confirm that, for any kind of favorable integer $n: (a + b)^n \geq a^n + b^n$ for all $a, b > 0:$

I threw away $3$ web pages of note pad paper on this trouble, and also I'm obtaining no place gradually. So I require some tips.

$1.$ What strategy would certainly you make use of to confirm this (e.g. induction, straight, counter instance)

$2.$ Are there any kind of methods to the evidence? I've seen some insane things took out of no place when it involves evidence ...

This adheres to straight from the binomial theorem. Conversely, you can confirm it inductively (which is possibly extra enjoyable) : intend the inequality real for $n-1$. After that $(a+b)^n = (a+b)(a+b)^{n-1} \geq (a+b)(a^{n-1} + b^{n-1})$ by the inductive theory. So $(a+b)^n \geq a(a^{n-1}+ b^{n-1}) + b(b^{n-1} + a^{n-1})$, and also this goes to the very least $a^n + b^n$.

It could additionally be handy for you to assume a little concerning the geometry of the inequality.

For $n=2$, locate a means to place an $a \times a$ square and also a $b \times b$ square right into a $(a+b) \times (a+b)$ square with no. overlaps. For $n=3$, see if you can fit an $a \times a \times a$ dice and also a $b \times b \times b$ dice within an $(a+b) \times (a+b) \times (a+b)$ dice without overlaps.

Next, the idea of having greater than 3 measurements could appear a little unusual, yet. consider package in $n$ measurements whose sides have size $a+b$. Can you fit 2 boxes within it, one with side size $a$ and also one with side size $b$?

Induction.

For n= 1 it is trivially real

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

Consider instance n= k +1 (a+ b) ^ (k +1) = (a+ b) (a+ b) ^ k >= (a+ b) (a ^ k+ b ^ k) =a ^ (k +1) +b ^ (k +1) +abdominal muscle ^ k+ bachelor's degree ^ k >= a ^ (k +1) +b ^ (k +1) as called for

Let's have a precalculus solution : Consider the function of $a$ relying on the parameter $b$ that is $f_b(a)=(a+b)^n-a^n-b^n$. Its acquired about $a$ is $f'_b(a)=n((a+b)^{n-1}-a^{n-1})$ due to the fact that $b>0$ $f'_b(a,b)$ is nonnegative and also $f_b$ is a raising function of $a$ and also you can end.

Hint : Use the binomial theory.

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

Currently, keep in mind that every term in the 2nd amount declares; this is due to the fact that a, b, and also the binomial coefficients are all favorable. Consequently, (a+ b ) ^{n } = a ^{n } + b ^{n } + (amount of favorable terms ) >= a ^{n } + b ^{n }.

Here is an additional means to consider the inequality. Select the bigger of $a^n$ and also $b^n$ and also divide via by that amount. This lowers the trouble to revealing that $(1+r)^n \ge 1 + r^n$ for. some favorable actual number $r \le 1$. If $r < 1$, we have $r^n < r$, so $1 + r^n < 1 + r < (1+r)^n$. I leave the instance $r = 1$ (and also $n$ sensible and also $\ge 1$ ) to others.