gavrilos wrote: Find the largest possible value of $M$ for which $\frac{x}{1+\frac{yz}{x}}+\frac{y}{1+\frac{zx}{y}}+\frac{z}{1+\frac{xy}{z}}\geq M$ for all $x,y,z>0$ with $xy+yz+zx=1$ The largest M is $\frac{3}{\sqrt{3}+1}$.This is my solution: By C-B-S we have $\frac{x}{1+\frac{yz}{x}}+\frac{y}{1+\frac{zx}{y}}+\frac{z}{1+\frac{xy}{z}}=\frac{x^2}{x+yz}+\frac{y^2}{y+xz}+\frac{z^2}{z+xy}\ge \frac{(x+y+z)^2}{x+y+z+xy+yz+zx}=\frac{(x+y+z)^2}{x+y+z+1}$.The function $f(x)=\frac{x^2}{x+1}$ is increasing and $x+y+z\ge \sqrt{3}$,so $\frac{(x+y+z)^2}{x+y+z+1}\ge \frac{3}{\sqrt{3}+1}$. We have equality when $x=y=z=\frac{1}{\sqrt{3}}$.

Nice. Putting M1=3/(1+SQRT(3)) we should prove (after using Cauchy-Schwarz) that 1+3(x+y+z)<=x^2+y^2+z^2+sqrt(3) (x+y+z)^2. Since x^2+y^2+z^2>=xy+yz+xz=1 and x+y+z>=sqrt(3*(xy+yz+zx))=sqrt(3) ==> (x+y+z)^2*sqrt(3)>=3(x+y+z), we are done. If M>M1, putting x=y=z we get a contradiction. Thus, M=M1=3/(1+sqrt(3)).

Now, notice that $\sum_{cyc}\frac{x}{1+\frac{yz}{x}}=\sum_{cyc}\frac{x^2}{x+yz}\overset{\text{Engel C-S}}{\ge}\frac{\left(\sum_{cyc}x\right)^2}{\sum_{cyc}x+\sum_{cyc}xy}$ Furthermore since $\left(\sum_{cyc}x\right)^2\ge3\sum_{cyc}xy=3$, we have that $\frac{\left(\sum_{cyc}x\right)^2}{\sum_{cyc}x+\sum_{cyc}xy}=\frac{\left(\sum_{cyc}x\right)^2}{\sum_{cyc}x+1}\ge\frac{3}{\sum_{cyc}x+1}$ Thus we obtain $3\ge M\left(\sum_{cyc}x+1\right)\Longrightarrow 3-M\sum_{cyc}x\ge M$, thus since we wish to maximize the left hand side, we must have $3-\sqrt{3}M\ge M\Longrightarrow \frac{3}{\sqrt{3}+1}\ge M$. Therefore the largest $M$ is equal to $\frac{3}{\sqrt{3}+1}$ $\blacksquare$.

@below, last part is wrong .