\begin{align*} \text{By AM-HM}\\ \sum_{cyc} \frac{x+2y}{z+2x+3y} \leq \sum_{cyc} (\frac{x+2y}{4})(\frac{1}{x+y+z}+\frac{1}{x+2y})=\frac{3}{2}\\ \end{align*}

$\frac{x + 2y}{z + 2x + 3y}+\frac{y + 2z}{x + 2y + 3z}+\frac{z + 2x}{y + 2z + 3x} =3-\frac{x+y+z}{z+2x+3y}-\frac{x+y+z}{x+2y+3z}-\frac{x+y+z}{y+2z+3x}=3-(x+y+z)(\frac{1}{x+2y+3z}+\frac{1}{y+2z+3x}+\frac{1}{z+2x+3y}) \leq 3-(x+y+z)\frac{9}{6(x+y+z)}=\frac{3}{2}$

parmenides51 wrote: Let $x, y, z$ be positive real numbers. Prove that: $\frac{x + 2y}{z + 2x + 3y}+\frac{y + 2z}{x + 2y + 3z}+\frac{z + 2x}{y + 2z + 3x} \le \frac{3}{2}$ : https://artofproblemsolving.com/community/c6h1087578p4816915 https://artofproblemsolving.com/community/c6h1087985p4821224 $$ \frac{6}{7}<\frac{x + 2y}{z + 2x + 3y}+\frac{y + 2z}{x + 2y + 3z}+\frac{z + 2x}{y + 2z + 3x} \le \frac{3}{2}$$

sqing wrote: parmenides51 wrote: Let $x, y, z$ be positive real numbers. Prove that: $\frac{x + 2y}{z + 2x + 3y}+\frac{y + 2z}{x + 2y + 3z}+\frac{z + 2x}{y + 2z + 3x} \le \frac{3}{2}$ : https://artofproblemsolving.com/community/c6h1087578p4816915 https://artofproblemsolving.com/community/c6h1087985p4821224 $$ \frac{6}{7}<\frac{x + 2y}{z + 2x + 3y}+\frac{y + 2z}{x + 2y + 3z}+\frac{z + 2x}{y + 2z + 3x} \le \frac{3}{2}$$ Multiply $(z+2x+3y)(x+2y+3z)(y+2z+3x)$, we get $\frac{6}{7}(6(\sum_{cyc}x^3)+23(\sum_{cyc}x^2y)+25(\sum_{cyc}x^2z)+54xyz)$ $<$ $7(\sum_{cyc}x^3)+33(\sum_{cyc}x^2y)+39(\sum_{cyc}x^2z)+87xyz$ Obviously this holds.And $\frac{3}{2}(6(\sum_{cyc}x^3)+23(\sum_{cyc}x^2y)+25(\sum_{cyc}x^2z)+54xyz)$ $\geq$ $7(\sum_{cyc}x^3)+33(\sum_{cyc}x^2y)+39(\sum_{cyc}x^2z)+87xyz$ $\Leftrightarrow$ $3(\sum_{cyc}x^2z)+12xyz$ $\leq$ $4(\sum_{cyc}x^3)+3(\sum_{cyc}x^2y)$ This holds by AM-GM inequality

parmenides51 wrote: Let $x, y, z$ be positive real numbers. Prove that: $\frac{x + 2y}{z + 2x + 3y}+\frac{y + 2z}{x + 2y + 3z}+\frac{z + 2x}{y + 2z + 3x} \le \frac{3}{2}$ : We can rewrite this inequality as $\frac{1}{z + 2x + 3y}+\frac{1}{x + 2y + 3z}+\frac{1}{y + 2z + 3x} \ge \frac{3}{2(x+y+z)}$ wichi true by C-S

Alternative solution: Look that $\sum_{cyc}{\frac{x+2y}{z+2x+3y}}=\sum_{cyc}{\frac{\frac{x+2y}{x+y+z}}{1+\frac{x+2y}{x+y+z}}}$, then we define that $f(a)=\frac{a}{1+a}$ wich is concave, then: $\frac{1}{3}\sum_{cyc}{f\left(\frac{x+2y}{x+y+z}\right)}\leq f\left(\frac{1}{3}\sum_{cyc}{\frac{x+2y}{x+y+z}}\right)=f(1)=\frac{1}{2}\Rightarrow \sum_{cyc}{\frac{\frac{x+2y}{x+y+z}}{1+\frac{x+2y}{x+y+z}}}\leq \frac{3}{2} \blacksquare$

Let's say x+2y=a y+2z=b z+2x=c so, 3/2>=a/a+b+b/b+c+c/c+a then use Tchebyshevs inequality 3(ab+ac+bc)>=(a/a+b +b/b+c +c/c+a)*(b(a+b)+c(b+c)+a(c+a) then inequality comes 3(a²+b²+c²+ab+ac+bc)>=6(ab+ac+bc) a²+b²+c²>=ab+ac+bc which is desired.