$(a+b)^n < 2^n(a^n + b^n)$ when $ a,b>0$ will it be possible for all natural number .
How to proceed with it by Mathematical Induction? can anyone help me?
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1$(a+b)^{n+1} = (a+b)^n(a+b) < 2^n(a^n + b^n)(a+b) = 2^n(a^{n+1} + ab^n + ba^n + b^{n+1}) = 2^n(a^{n+1} + b^{n+1) + 2^n(ab^n + ba^n)$. Can you proof that $ab^n + ba^n < a^{n+1} + b^{n+1}$? Maybe be AM-GM? – fleablood Jan 07 '18 at 07:47
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1Actually, the stronger $,(a+b)^n \le 2^{n\color{red}{-1}}(a^n + b^n) \iff \displaystyle \frac{a+b}{2} \le \sqrt[n]{\frac{a^n + b^n}{2}},$ holds true by the generalized means inequality. – dxiv Jan 07 '18 at 07:56
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Without loss of generality, suppose $a \geq b$.
Then, noting that $a, b > 0$, we have the following for $n > 0$:
$$(a+b)^n \leq (a+a)^n = (2a)^n = 2^n \cdot a^n < 2^n \cdot a^n + 2^n \cdot b^n = 2^n (a^n + b^n)$$
as desired.
Benjamin Dickman
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1@Taroccoesbrocco True! Okay, rather than edit, here is a full proof for the $n=0$ case: $1 < 2$. Q.E.D. $$ $$ (But actually: Thanks!) – Benjamin Dickman Jan 07 '18 at 07:55
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@PeterSzilas I assume you refer to the $n=0$ proof in the comments: Thank you. – Benjamin Dickman Jan 07 '18 at 07:57
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@cmi You decided one month later to un-accept this as the answer, down-vote, and call it "obvious"? – Benjamin Dickman Feb 07 '18 at 12:26
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It is true for $n=1$: $$(a+b)<2(a+b)$$ Assuming it is true for $n=k$, we'll show it's true for $n=k+1$: $$(a+b)^{k+1}=(a+b)^k(a+b)<2^k(a^k+b^k)(a+b)=2^k(a^{k+1}+b^{k+1})+2^k(a^kb+ab^k)<2^{k+1}(a^{k+1}+b^{k+1}),$$ because due to rearrangement: $$a^kb+ab^k<a^{k+1}+b^{k+1}.$$
farruhota
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it is the rearrangement inequality. you can check this out: https://en.wikipedia.org/wiki/Rearrangement_inequality – farruhota Feb 07 '18 at 08:09
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