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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?

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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.

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