An inequality with real index

Question

I have tried this problem but not able to find the answer. Could anybody help me in solving this. Is $(a+b)^p< a^p+b^p$ for all non-negative reals $a, b$ and $0<p<1?$

Also of Prove that $(p+q)^m \leq p^m+q^m$ and a gazillion of other questions. Search... — , Oct 13 '15 at 13:15

score 3 · Answer 1 · answered Sep 23 '15 at 06:27

$\forall 0 < x, y < 1 , \,\, x^y > x.$ Thus

$$ \forall a,b, > 0,\,\, 0 < p < 1, \,\, \left(\frac{a}{a + b}\right)^p + \left(\frac{b}{a + b}\right)^p > \left(\frac{a}{a + b}\right) + \left(\frac{b}{a + b}\right) = 1$$

If $a = 0$ or $b = 0$ there is equality.

score 0 · Answer 2 · answered Sep 23 '15 at 08:38

0

I got another idea. The function $x^{p}$ is sub-additive in this case. So the result follows directly.

answered Sep 23 '15 at 08:38

user159888

1,988

And how do you prove that $x \mapsto x^p$ is sub-additive? – Siminore Sep 23 '15 at 08:39
The second derivative of $x^p$ is negative for $0<p<1$ and hence $x^p$ is concave. Every concave function is sub-additive. Proving this part is easy as you can refer https://en.wikipedia.org/wiki/Concave_function. – user159888 Sep 23 '15 at 09:19

An inequality with real index

2 Answers2