$(x+y)^r \le x^r+y^r$ when $r \in (0,1)$ and $x,y$ are real positive numbers.

Asked May 21 '19 at 14:08

Active May 21 '19 at 14:13

Viewed 162 times

I'm sorry for the silly question, but I have a doubt. Given two positive real numbers $x,y$ and taking $0<r<1$, is it true that

$$ (x+y)^r \le x^r+y^r? $$

In all the examples I considered, this turns out to be true. Moreover, to try to give a proof, does it suffices to prove that the function $f(z)=(1+z)^r-1-z^r$ is non-positive when $z \in (0,1]$ (i.e. taking $x>y$ and dividing by $x$?

asked May 21 '19 at 14:08

TheWanderer

5,378

3

Possible duplicate of $(x + y)^s \le x^s + y^s$ for $0 < s \le 1$ – Martin R May 21 '19 at 14:13
Also: https://math.stackexchange.com/q/54964, https://math.stackexchange.com/q/1155225/42969, https://math.stackexchange.com/q/233146, https://math.stackexchange.com/q/49210, https://math.stackexchange.com/q/264156 – Martin R May 21 '19 at 14:14
You can read this: https://en.wikipedia.org/wiki/Binomial_theorem#Newton's_generalized_binomial_theorem – Anas Khaled May 21 '19 at 21:09

$(x+y)^r \le x^r+y^r$ when $r \in (0,1)$ and $x,y$ are real positive numbers.

0 Answers0