Prove that $ (x+y)^n \le x^n +y^n$ with $x,y \ge 0 $ and $0 \lt n \le 1$

Question

I have no clue on this one so I hope you can help me out on this one.

Let $x,y \ge 0 $ and $0 \lt n \le 1$. Then $$ (x+y)^n \le x^n +y^n$$

Answer 1

Since our inequality is symmetric, we can assume that $y\geq x$.

Also if $x=0$ so the inequality is obviously true.

Let $x>0$ and $y=tx$, where $t\geq1$.

Hence, we need to prove that $f(t)\geq0$, where $f(t)=t^n+1-(t+1)^n$.

But, $f'(t)=n\left(\frac{1}{t^{1-n}}-\frac{1}{(1+t)^{1-n}}\right)\geq0$.

Thus, $f(t)\geq f(1)=0$ and we are done!