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 ...

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2019-05-04 16:23:04
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Answers: 6

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 $a^n + b^n$.

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2019-05-08 01:41:09
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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 $b$?

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2019-05-08 01:39:56
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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

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2019-05-08 01:34:27
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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.

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2019-05-08 01:28:58
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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 .

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2019-05-07 23:32:13
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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.

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2019-05-07 23:29:49
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