RaleD wrote: Let $a, b, c$ be positive reals such $a+b+c=1$. Prove: $a \sqrt[3]{1+b-c} + b\sqrt[3]{1+c-a} + c\sqrt[3]{1+a-b} \leq 1$ Trivial, indeed.

By AM-GM: $3a\sqrt[3]{1+b-c} \leqslant a(1+b-c +1 +1)$. Adding the other similar inequalities we have ${3(a\sqrt[3]{1+b-c}}+b\sqrt[3]{1+c-a} +c\sqrt[3]{1+a-b} \leqslant 3(a+b+c)=3$, hence done.

Posted not long ago, from Japan anyway: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=197379&sid=2f5de50b93026cd22d97ec742bb45e92#p197379

Let $f(x)=\sqrt[3]{x}=x^{\frac{1}{3}}, f'(x)=\frac{1}{3}x^{-\frac{4}{3}}\text{ and }f''(x)=-\frac{4}{9}x^{-\frac{7}{3}}<0, \forall x\in \mathbb{R}^+$ therefore the function is concave for all $x$ in positive reals. $\sum_{cyc}a\sqrt[3]{1+b-c}=\sum_{cyc}af(x_1)\overset{\text{W-Jensen}}{\le}f(ax_1+bx_2+cx_3)=f\left(\sum_{cyc}a(1+b-c)\right)=f\left(\sum_{cyc}a\right)=f(1)$ Thus $\sum_{cyc}a\sqrt[3]{1+b-c}\le f(1)=\sqrt[3]{1}=1$ $\blacksquare$

RaleD wrote: Let $a, b, c$ be positive reals such that $a+b+c=1$. Prove that the inequality \[a \sqrt[3]{1+b-c} + b\sqrt[3]{1+c-a} + c\sqrt[3]{1+a-b} \leq 1\]holds. Japan 2005 https://artofproblemsolving.com/community/c4h524818p3353030

Use holder inequality we have $T^3=(\sqrt[3]{a}\sqrt[3]{a}\sqrt[3]{a+ab-ac}+\sqrt[3]{b}\sqrt[3]{b}\sqrt[3]{b+cb-ab}+\sqrt[3]{c}\sqrt[3]{c}\sqrt[3]{c+ac-bc})\le (a+b+c)^3=1$ Q.E.D