JBMO2020 wrote: Let $a, b, c$ be non-negative reals which satisfy $a+b+c=1$. Prove that $\frac{\sqrt{a}}{b+1}+\frac{\sqrt{b}}{c+1}+\frac{\sqrt{c}}{a+1}>\frac{1}{2}(\sqrt{a}+\sqrt{b}+\sqrt{c})$ Do u mean this?

By AM-GM and Cauchy-Schwarz's inequality and Power-mean, we have \begin{align*}\sum_{cyc}\frac{\sqrt{a}}{b+1}&=\sum_{cyc}\left(\sqrt{a}-\frac{b\sqrt{a}}{b+1}\right) \\&\ge \sum_{cyc}\left(\sqrt{a}-\frac{\sqrt{ab}}{2}\right) \\&\ge \sum_{cyc}\sqrt{a}-\frac{1}{6}\left(\sum_{cyc}\sqrt{a}\right)^2 \\&\ge \sum_{cyc}\sqrt{a}-\frac{1}{2\sqrt{3}}\sum_{cyc}\sqrt{a} \\&> \frac{1}{2}\sum_{cyc}\sqrt{a}\end{align*} Note: The following inequality is also true with the same condition. $$\sum_{cyc}\frac{\sqrt{a}}{b+1} \ge \frac{3}{4}\sum_{cyc}\sqrt{a}$$I will post my proof later. If someone can solve it, please feel free to post your solution.

@above: how can you do that? $\sqrt{a}-\frac{b\sqrt{a}}{b+1} \geq \sqrt{a}-\frac{\sqrt{ab}}{2}$ - First: $a+b+c=1$ so $b+1 \geq 2\sqrt{b}$ is a big problem - Second: $a;b;c$ is non-nagative, not positive, that's meaning $-\frac{b\sqrt{a}}{b+1} \geq -\frac{b\sqrt{a}}{2\sqrt{b}}$ can be division by zero, another mistake.

@above, why $b+1\ge 2\sqrt{b}$ is a big problem? Secondly, $-\frac{b\sqrt{a}}{b+1} \ge -\frac{\sqrt{ab}}{2}$ is still true for $a,b,c \ge 0$.

@above: Sorry, my bad, you're right

NaPrai wrote: Note: The following inequality is also true with the same condition. $$\sum_{cyc}\frac{\sqrt{a}}{b+1} \ge \frac{3}{4}\sum_{cyc}\sqrt{a}$$I will post my proof later. If someone can solve it, please feel free to post your solution. Have you a nice proof?

arqady wrote: Have you a nice proof? Yes, I have a nice proof.