Thread: How can this inequality be proved?

1. How can this inequality be proved?

If x, y > 0 and 0 < p < 1, then (x+y)^p < x^p+y^p.
It was on Wikipedia:
http://en.wikipedia.org/wiki/Inequality_(mathematics)#Power_inequalities

2. Re: How can this inequality be proved?

Originally Posted by Karima
If x, y > 0 and 0 < p < 1, then (x+y)^p < x^p+y^p.
It was on Wikipedia:
http://en.wikipedia.org/wiki/Inequality_(mathematics)#Power_inequalities
There are a few fun ways to prove this.

First, consider the case
$\, x=y$.
Note then that
$(x+y)^p = (x+x)^p = (2x)^p = 2^p(x^p)$
and
$x^p + y^p = x^p + x^p = 2(x^p)$.
Since
$0,
it follows that
$2^p<2$,
and therefore,
$(x+y)^p < x^p+y^p$ when $x=y$.

Next, assume, Without Loss of Generality, that $\, x>y$.

$(x+y)^2 = \left(x \left(1 + \tfrac{y}{x} \right)\right)^p = x^p \left(1+\tfrac{y}{x}\right)^p$

Since $\tfrac{y}{x}<1$, we have an expansion of the form
$(1+r)^p$, $0.

I'm a bit tired, so perhaps you can finish this or I will later on.

3. Re: How can this inequality be proved?

Let me rephrase your inequality so what follows is more clear: $(a+b)^p where $a,b>0$ and $0.
If $a=b$, $2^pa^p<2a^p$ since $2^p<2$ -- the last inequality since the function $x^p$ is increasing.
So suppose $a\ne b$. I can suppose a<b. (The problem is symmetric in a and b.) Consider the function $f(x)=(x+1)^p-x^p$ on $(0,\infty)$. Now $f^{\prime}(x)<0$ --
$(x+1)^{p-1}-x^{p-1}<0$ since $x^{p-1}$ is decreasing. Since $\lim_{x\rightarrow 0^+}f(x)=1$, the function f can be extended to a continuous function $g$ on $[0,\infty)$ where $g(0)=1$ and $g(x)=f(x)$ otherwise. Now $g$ is decreasing on $[0,\infty)$ -- the proof that a function on $[0,\infty)$, which has a negative derivative in the open interval $(0,\infty)$, uses the mean value theorem on $[0,\infty)$; remember that differentiability is required only on the open interval. So for any $x>0, 1=g(0)>g(x)=(x+1)^p-x^p$ for any $x>0$ or $(x+1)^p
Finally, $(a/b+1)^p<(a/b)^p+1$ and easily then $(a+b)^p.