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Why does this inequality hold?

$$(x+y)^\alpha \leq x^\alpha + y^\alpha$$

where $x$, $y\geq 0$ and $\alpha \in (0,1)$.

Thanks in advance!

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    $\begingroup$ Also, you must mean $x,y\geq 0$ for it to make sense, right? $\endgroup$ – String Apr 8 '15 at 7:47
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    $\begingroup$ Fix $y\ge 0.$ Then you have two functions of $x$ that agree at $x=0.$ Compare derivatives. $\endgroup$ – zhw. Apr 8 '15 at 7:54
  • $\begingroup$ Dividing by $y^\alpha$, and letting $X=\frac xy$, we see that we only need to prove that $(X+1)^\alpha\le X^\alpha+1$. $\endgroup$ – Akiva Weinberger Apr 8 '15 at 8:16
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NOTE: This was an attempt to make the proof as elementary as possible, so I deliberately avoided using calculus and the like.


Fix $\alpha\in(0,1)$ and consider the function $$ f(t)=t^\alpha,\quad t\geq 0 $$ Then note that $f(kt)=k^\alpha t^\alpha$. For $k>1$ we have $k^\alpha<k^1=k$ since $s\mapsto k^s$ is a strictly increasing function. Thus we get the relation $$ f(kt)=k^\alpha f(t)<k f(t),\quad\forall k>1 $$


So we will use this to show that $$ f(x+y)\leq f(x)+f(y) $$ This clearly holds with equality if either $x=0$ or $y=0$. And if $x=y>0$ we have $f(x+y)=f(2x)<2f(x)=f(x+f(y)$. So assume WLOG that $0<x<y$. Then $k=\frac yx>1$ so that $$ f(y)=f\left(\frac yx\cdot x\right)<\frac yx f(x)\iff f(x)>\frac xy f(y) $$ and thus $$ \begin{align} f(x+y)&=f\left(\left(\frac xy+1\right)y\right)\\ &<\left(\frac xy+1\right)f(y)\\ &<f(x)+f(y) \end{align} $$ which finishes the last part of what we wanted to show.

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Hint: each site is not smaller than zero. Thus look at the power of $1/\alpha>1$

$$x+y\leq \left(x^\alpha + y^\alpha\right)^{1/\alpha} = x+y+a$$

with $a\ge 0$

There are various methods to estimate $a$ For instance using the http://en.wikipedia.org/wiki/Binomial_theorem with $\beta = 1/\alpha>1$ $$(x+y)^{\alpha} =x^\beta\left(1+\tfrac{y}{x}\right)^\alpha =x^\beta\sum_{k=0}^{\infty}\binom{\beta}{k}\left(\frac{y}{x}\right)^k =\sum_{k=0}^{\infty}\binom{\beta}{k}x^{\beta - k}y^{k}$$ Then $$\left(x^\alpha + y^\alpha\right)^{\beta}=x^{\alpha\cdot\beta}+y^{\alpha \text{ floor}(\beta) }+\sum_{k=1}^{\text{ floor}(\beta)}\binom{\beta}{k}x^{(\beta - k)\cdot\alpha}y^{k\cdot\alpha}$$ With $\text{ floor}(\beta)\cdot\alpha\le 1$ and $y\le1$ $y^{\alpha \text{ floor}(\beta) }\ge y$ Thus we get: $$\left(x^\alpha + y^\alpha\right)^{\beta}\ge x+y+\sum_{k=1}^{\text{ floor}(\beta)}\binom{\beta}{k}x^{(\beta - k)\cdot\alpha}y^{k\cdot\alpha}$$

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Let $f(x) = (x+y)^\alpha - x^\alpha - y^\alpha$.

We have to show that $f(x) \leq 0, \forall x \geq 0$

We get $f'(x) = \alpha (x + y)^{\alpha - 1} - \alpha x^{\alpha - 1} = \alpha( (x + y)^{\alpha - 1} - x^{\alpha - 1})$. Notice that $\alpha - 1 \in (-1,0)$. So since $x + y \geq x$, $(x+y)^{\alpha - 1} \leq x^{\alpha - 1}$, we get $f'(x) \leq 0$.

$f(0) = y^\alpha - y^\alpha = 0$

So the result we wanted to prove follows.

[Edit] Fixed line 4.

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  • $\begingroup$ 4th line: exponent should be $a-1.$ $\endgroup$ – zhw. Apr 8 '15 at 8:25

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