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Let $x,y>0$ and $p>1$. Is it always true that $$(x+y)^p>x^p+y^p?$$

I think it is true. But can not prove it.

Marti
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    Did you even try? Expand $(x+y)^p$ using the Binomial Theorem... – Hubble Oct 04 '15 at 20:56
  • https://math.stackexchange.com/questions/1368, https://math.stackexchange.com/questions/264156 – sdcvvc Oct 04 '15 at 20:57
  • Expanding for fractional $p$ would not yield any negative coefficients...? – Marti Oct 04 '15 at 20:57
  • How? You specified $x,y>0$. – Hubble Oct 04 '15 at 20:59
  • Thanks@sdcvvc. So, it is true..? – Marti Oct 04 '15 at 20:59
  • If you can't answer this question after my comment and the links to two other questions asking the same thing, then I don't know what to say. – Hubble Oct 04 '15 at 21:00
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    @iHubble How can the OP expand $(x+y)^p$ using the Binomial Theorem if $p$ is not an integer? – thanasissdr Oct 04 '15 at 21:02
  • Hi guys, thanks for all the comments and help. sdcvvc's answer is great and it is true. – Marti Oct 04 '15 at 21:03
  • @thanasissdr, great question. I immediately assumed $p \in \mathbb{N}$ since this letter is usually reserved for integers (e.g. $p$-adic numbers, $p$-norms, etc.) – Hubble Oct 04 '15 at 21:05
  • @iHubble I think the OP should state if $p>1$ is either an integer or any real number, otherwise I don't know if the Binomial Theorem would be helpful, at least at this form of the problem. – thanasissdr Oct 04 '15 at 21:08
  • @thanasissdr, there is a generalized version of the Binomial Theorem that holds for any exponent $r \in \mathbb{C}$. See here. – Hubble Oct 04 '15 at 21:10
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    removed my answer which was indeed a comment - and here is the comment: For the case that $x,y\geq 0$ it is, otherwise you could choose $x=-y$ and get a contradiction. – Max Oct 04 '15 at 21:12
  • @iHubble Sure, but things will get pretty complicated (infinite sum).. – thanasissdr Oct 04 '15 at 21:17

4 Answers4

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Since the inequality is homogeneous, we may assume without loss of generality that $y=1$ and $x\in(0,1]$, then prove: $$ \forall x\in(0,1],\qquad (1+x)^p > 1+x^p.\tag{1}$$ On the other hand, if we set $f(x)=(1+x)^p-x^p$, it is straightforward to check that $f'(x) = (p-1)\cdot\left((1+x)^{p-1}-x^{p-1}\right)>0$, hence $f$ is increasing on $(0,1]$. Since $f(0)=1$, $(1)$ is proved.

Jack D'Aurizio
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    This is a new approach, while I found answers by timon92 and A1DHTH are essentially same as that by sdcvcc (look at the link by his/her comments). I upvoted both the answers. – Marti Oct 04 '15 at 21:27
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Obviously for any $1>t>0$ and $p>1$ we have $t>t^p$ thus $$\frac{x}{x+y} > \left( \frac{x}{x+y} \right)^p \hbox{ and } \frac{y}{x+y} > \left( \frac{y}{x+y} \right)^p.$$ Summing up and multiplying by $(x+y)^p$ we get $(x+y)^p > x^p+y^p$.

timon92
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Since $0 < \dfrac{x}{x+y} < 1, 0 < \dfrac{y}{x+y} < 1 \Rightarrow \left(\dfrac{x}{x+y}\right)^p < \dfrac{x}{x+y}, \left(\dfrac{y}{x+y}\right)^p < \dfrac{y}{x+y}$. Adding these inequalities, the answer follows.

DeepSea
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Let $x,y>0$ and $p>1$ so we must prove $(x+y)^p > x^p + y^p$. We can prove this by saying that $(x+y)^p = x^p + y^p + c(x,y)$ where $c(x,y)$ are terms of $x,y$ that are combined. If $x,y>0$ then so must $c(x,y)$ because it will be the sum of products of positive numbers. Example for $c(x,y)$ $(x+y)^2 = x^2 + y^2 + 2xy$. Thus as we know $c(x,y)>0$ we can say $(x+y)^p = x^p + y^p + c(x,y) > x^p + y^p$ and if we subtract $x^p + y^p$ from both sides we get $c(x,y)>0$ which is true.

Eli Sadoff
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