By AM-GM $$\frac{\sqrt{ab}}{a+b+2c} \leqslant \frac{\sqrt{ab}}{2\sqrt{(a+c)(b+c)}} \leqslant \frac{\frac{a}{a+c}+\frac{b}{b+c}}{4}.$$Therefore, we have $$\sum_\text{cyc} \frac{\sqrt{ab}}{a+b+2c} \leqslant \sum_\text{cyc} \frac{\frac{a}{a+c}+\frac{b}{b+c}}{4} = \frac{3}{4}.$$

MilosMilicev wrote: Show that for $a,b,c > 0$ the following inequality holds: $$\frac{\sqrt{ab}}{a+b+2c}+\frac{\sqrt{bc}}{b+c+2a}+\frac{\sqrt{ca}}{c+a+2b} \le \frac {3}{4}$$. Let $a$, $b$ and $c$ be positive numbers. Prove that:\[ \frac{\sqrt{ab}}{a+b+2c}+\frac{\sqrt{bc}}{b+c+2a}+\frac{\sqrt{ca}}{c+a+2b} \leq\frac{a}{2a+b+c }+\frac{b}{2b+ c+a}+\frac{c}{2c+a+b }\le \frac {3}{4}\] here

MilosMilicev wrote: Show that for $a,b,c > 0$ the following inequality holds: $\frac{\sqrt{ab}}{a+b+2c}+\frac{\sqrt{bc}}{b+c+2a}+\frac{\sqrt{ca}}{c+a+2b} \le \frac {3}{4}$. But you do this: $x,y,z>0$,prove \[\frac{3}{5}\geq {\frac {\sqrt {yz}}{z+3\,x+y}}+{\frac {\sqrt {zx}}{x+3\,y+z}}+{ \frac {\sqrt {xy}}{y+3\,z+x}}\] \[{\frac {\sqrt {yz}}{z+3\,x+y}}+{\frac {\sqrt {zx}}{x+3\,y+z}}+{\frac { \sqrt {xy}}{y+3\,z+x}}\geq \frac{4}{5}(\,{\frac {\sqrt {yz}}{2\,x+z+y}}+{ \frac {\sqrt {zx}}{2\,y+x+z}}+{\frac {\sqrt {xy}}{2\,z+y+x}})\]

xzlbq wrote: But you do this: $x,y,z>0$,prove \[\frac{3}{5}\geq {\frac {\sqrt {yz}}{z+3\,x+y}}+{\frac {\sqrt {zx}}{x+3\,y+z}}+{ \frac {\sqrt {xy}}{y+3\,z+x}}\] https://artofproblemsolving.com/community/c6h470757p2636145

MilosMilicev wrote: Show that for $a,b,c > 0$ the following inequality holds: $$\frac{\sqrt{ab}}{a+b+2c}+\frac{\sqrt{bc}}{b+c+2a}+\frac{\sqrt{ca}}{c+a+2b} \le \frac {3}{4}$$ Show that for $a,b,c,d > 0$ the following inequality holds: $$\frac{\sqrt[3]{abc}}{a+b+c+3d}+\frac{\sqrt[3]{bcd}}{b+c+d+3a}+\frac {\sqrt[3]{cda}}{c+d+a+3b}+\frac{\sqrt[3]{dab}}{d+a+b+3c} \le \frac {2}{3}.$$h

sqing wrote: Show that for $a,b,c,d > 0$ the following inequality holds: $$\frac{\sqrt[3]{abc}}{a+b+c+3d}+\frac{\sqrt[3]{bcd}}{b+c+d+3a}+\frac {\sqrt[3]{cda}}{c+d+a+3b}+\frac{\sqrt[3]{dab}}{d+a+b+3c} \le \frac {2}{3}.$$h Here we manage to show the following general result: Let $S=\sum_{k=1}^nx_k$ , and $P_i=\frac{1}{x_i}\displaystyle\prod_{k=1}^nx_k$ , $i\in\{1,\cdots,n\}$, $n\ge2$, where $x_i$ are positive real numbers. It holds$$\sum_{i=1}^n\frac{\sqrt[n-1]{P_i}}{S+(n-2)x_i}\le\frac{n}{2(n-1)}.$$Solution. Applying the AM-GM Inequality we obtain that \begin{align*}\sum_{i=1}^n\frac{\sqrt[n-1]{P_i}}{S+(n-2)x_i}=&\sum_{i=1}^n\frac{\sqrt[n-1]{P_i}}{\sum_{k\ne i}(x_i+x_k)}\\ \le&\sum_{i=1}^n\frac{\sqrt[n-1]{P_i}}{(n-1)\sqrt[n-1]{\prod_{k\ne i}(x_i+x_k)}}\\ =&\frac{1}{n-1}\sum_{i=1}^n\left(\prod_{k\ne i}\sqrt[n-1]{\frac{x_k}{x_i+x_k}}\right)\\ \le&\frac{1}{(n-1)^2}\sum_{i=1}^n\left(\sum_{k\ne i}\frac{x_k}{x_i+x_k}\right)\\ =&\frac{C_n^2\cdot1}{(n-1)^2}=\frac{n}{2(n-1)}, \end{align*}where $C_n^2=\frac{n!}{2!(n-2)!}=\frac{(n-1)n}{2}$ , and $\sum_{k\ne i}$ and $\prod_{k\ne i}$ denotes summing and multiplying over $k\in\{1,\cdots,n\}\setminus\{i\}$. $\blacksquare$

MilosMilicev wrote: Show that for $a,b,c > 0$ the following inequality holds: $\frac{\sqrt{ab}}{a+b+2c}+\frac{\sqrt{bc}}{b+c+2a}+\frac{\sqrt{ca}}{c+a+2b} \le \frac {3}{4}$. Stronger is: $x,y,z>0$,prove that \[\frac{3}{2}\,\sqrt {2}\geq {\frac {\sqrt {6\,yz+{y}^{2}+{z}^{2}}}{z+2\,x+y}}+{ \frac {\sqrt {6\,xz+{z}^{2}+{x}^{2}}}{2\,y+z+x}}+{\frac {\sqrt {6\,xy+ {x}^{2}+{y}^{2}}}{2\,z+x+y}}\]

xzlbq wrote: MilosMilicev wrote: Show that for $a,b,c > 0$ the following inequality holds: $\frac{\sqrt{ab}}{a+b+2c}+\frac{\sqrt{bc}}{b+c+2a}+\frac{\sqrt{ca}}{c+a+2b} \le \frac {3}{4}$. Stronger is: $x,y,z>0$,prove that \[\frac{3}{2}\,\sqrt {2}\geq {\frac {\sqrt {6\,yz+{y}^{2}+{z}^{2}}}{z+2\,x+y}}+{ \frac {\sqrt {6\,xz+{z}^{2}+{x}^{2}}}{2\,y+z+x}}+{\frac {\sqrt {6\,xy+ {x}^{2}+{y}^{2}}}{2\,z+x+y}}\] I try do it,and FAIL