# $(a+b)^\beta \leq a^\beta +b^\beta$ for $a,b\geq0$ and $0\leq\beta\leq1$ [duplicate]

It seems that $(a+b)^\beta \leq a^\beta +b^\beta$ for $a,b\geq0$ and $0\leq\beta\leq1$. However, I could not prove this nor the same result for a general concave and increasing function (for which it might not hold). If the inequality is true, does it follow from some general inequality or is there some other simple proof?

When $a=b=0$, the result is obvious. Assume that $(a,b)\ne(0,0)$. Then, $t=a/(a+b)$ is in $[0,1]$ hence $t^{\beta}\geqslant t$ because $\beta\leqslant1$. Likewise, $1-t=b/(a+b)$ is in $[0,1]$ hence $(1-t)^{\beta}\geqslant 1-t$. Summing these two inequalities yields $t^{\beta}+(1-t)^{\beta}\geqslant 1$, which is your result.