# $(x + y)^s \le x^s + y^s$ for $0 < s \le 1$ [duplicate]

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I'm reading a book on valuations on a field(Hasse's number theory). He says $(x + y)^s \le x^s + y^s$ for $0 < s \le 1$ and $x, y \ge 0$. I tried to prove it but failed.

Another question: Is $(x + y)^s \ge x^s + y^s$? where, $s \gt 1$ and $x, y \ge 0$.

OK, the first questions is duplicate, but IS THE SECOND QUESTION DUPLICATE, TOO?

## marked as duplicate by user147263, Daniel W. Farlow, Macavity, Claude Leibovici, drhabFeb 28 '15 at 8:47

Observe that for $0\le a\le 1$ and $0<s\le 1$, we have $a^{1-s}\le 1 \implies a\le a^s$
Note that if $x=y=0$ the inequality is trivially true. For $x+y>0$, using the observation we have $$\left(\frac{x}{x+y}\right)^s+\left(\frac{y}{x+y}\right)^s \ge \frac{x}{x+y}+\frac{y}{x+y}=1$$ $$\implies x^s+y^s \ge (x+y)^s$$
If $y=0$ then we have equality. If $y>0$ then devide by $y^s$ to get $(1+t)^s\ge 1+t^s$ Consider the function $f(x)=(1+t)^x-1-t^x$ Note $f'(x)=\log (1+t) (1+t)^x- t^x\log t>0$ since $(1+t)^x>t^x>0$ and $\log (1+t)>\log t$ we have that $f$ is increasing but $f(0)=0$ so $f(s)\ge 0$ for $s\ge 0$ which implies the inequality.