It equivalent to $$\sum_{cyc}\dfrac{1}{x^2+yz+1}\ge 1$$By T2's Lemma $$\sum_{cyc}\dfrac{1}{x^2+yz+1}\ge \dfrac{9}{x^2+y^2+z^2+xy+yz+zx+3}$$So it suffices to prove $9\ge x^2+y^2+z^2+xy+yz+zx+3\implies 3\ge xy+yz+zx\implies x^2+y^2+z^2\ge xy+yz+zx$ true.

sqing wrote: The positive real $x, y, z$ are such that $x^2+y^2+z^2 = 3$. Prove that$$\frac{x^2+yz}{x^2+yz +1}+\frac{y^2+zx}{y^2+zx+1}+\frac{z^2+xy}{z^2+xy+1}\leq 2$$ We have \[2-\sum{\frac{3(x^2+yz)}{4x^2+y^2+3yz+z^2}}=\frac{1}{2}\sum{\frac{(4x^2+3xy+4y^2)(x-y)^2+(x-y)^2(x+y-2z)^2}{(x^2+3xz+4y^2+z^2)(4x^2+y^2+3yz+z^2)}}\ge{0}\]

sqing wrote: The positive real $x, y, z$ are such that $x^2+y^2+z^2 = 3$. Prove that$$\frac{x^2+yz}{x^2+yz +1}+\frac{y^2+zx}{y^2+zx+1}+\frac{z^2+xy}{z^2+xy+1}\leq 2$$ $x^2+y^2+z^2\geq xy+yz+zx\implies \sum (x^2+yz+1)\leq 9$ By C-S, $(\sum (x^2+yz+1))\left(\sum\frac{1}{x^2+yz+1}\right)\geq (1+1+1)^2\implies \sum\frac{1}{x^2+yz+1}\geq 1$ $$\sum\frac{x^2+yz}{x^2+yz +1}=3-\sum\frac{1}{x^2+yz+1}\geq 2$$Q.E.D. $\blacksquare$

$$\sum\frac{x^2+yz}{x^2+yz +1}=3-\sum\frac{1}{x^2+yz+1}\leq 2$$

$\sum_{\text{cyc}}\frac{x^2+yz}{x^2+yz+1}=3-\sum_{\text{cyc}}\frac1{x^2+yz+1}\le3-\frac9{x^2+y^2+z^2+xy+yz+zx+3}\le3-\frac9{2(x^2+y^2+z^2)+3}=2$

We can convert the inequality into proving $\frac{1}{x^2+yz+1}+\frac{1}{y^2+xz+1}+\frac{1}{z^2+xy+1}\ge 1$, we can use AM-HM since all the terms on the left are clearly positive to yields $\frac{\frac{1}{x^2+yz+1}+\frac{1}{y^2+xz+1}+\frac{1}{z^2+xy+1}}{3}\ge\frac{3}{x^2+y^2+z^2+xy+zx+yz+3}$, since we are given $x^2+y^2+z^2=3$, this inequality becomes $\frac{\frac{1}{x^2+yz+1}+\frac{1}{y^2+xz+1}+\frac{1}{z^2+xy+1}}{3}\ge\frac{3}{6+xy+zx+yz}$. Now all we need to prove is $\frac{9}{6+xy+yz+xz}\ge 1$, To do this we utilize the well known fact $x^2+y^2+z^2\ge xy+yz+xz$, from this we can substitute what we are given to get $3\ge xy+yz+xz$, Adding $6$ to both sides yields $9\ge xy+yz+xz+6$, then dividing by 9 gives $1\ge \frac{xy+yz+xz}{9}$, We are allowed to take the recirprocal of both sides but we must flip the inequality to yield $\frac{9}{xy+yz+xz}\ge 1$ which is the desired inequality. Can someone check this solution

Applying Jensen on $f(x)=\frac{x}{x+1}$ (note that it is concave function) We have \begin{align*} \sum_{cyc}\frac{x^2+yz}{x^2+yz+1} \leq \ \sum_{cyc} \frac{2x^2}{2x+x^2} \rightarrow \sum_{cyc}\frac{x}{x+2} \end{align*}We have to prove $\sum_{cyc} \frac{x}{x+2} \leq 1$, simplifying we get \begin{align*} & \frac{3xyz+4(x+y+z)+4(xy+yz+zx)}{xyz+2(xy+xz+yz)+4(x+y+z)+8} \leq 1 \\& 3xyz+4(x+y+z)+4(xy+zy+yx) \leq xyz+4(xy+xz+yz)+4(x+y+z) \\ & 0 \leq 8-2xyz-2(xy+xz+yz) \\& 0 \leq 4-xyz-(xy+xz+yz) ,\text {because of the condition we have $x^2+y^2+z^2 \geq xy+yz+zx$ and $1 \geq xyz$}\\& 0 \leq 3-(xy+xz+yz) \\& 0 \leq 0& \end{align*}Equality occurs at $x=y=z=1$

Notice \begin{align*}\sum_{\text{cyc}}(x-y)^2\ge 0&\implies 2(x^2+y^2+z^2-xy-yz-zx)\ge 0\\&\implies 3=x^2+y^2+z^2\ge xy+yz+zx\end{align*}Hence, by Titu's Lemma, $$\sum_{\text{cyc}}\frac{1}{x^2+yz+1}\ge \frac{(1+1+1)^2}{3+(x^2+y^2+z^2)+(xy+yz+zx)}\ge 1.$$We see $\frac{1}{x^2+yz+1}=1-\frac{x^2+yz}{x^2+yz+1},$ which suffices. $\square$

$\frac{x^2+yz}{x^2+yz +1}+\frac{y^2+zx}{y^2+zx+1}+\frac{z^2+xy}{z^2+xy+1}=\sum\frac{x^2+yz}{x^2+yz +1}$ $=$ $=$$\sum\frac{x^2+yz+1-1}{x^2+yz +1}=\sum(\frac{x^2+yz+1}{x^2+yz +1}-\frac{1}{x^2+yz +1})$ $=$ $=$$\sum(1-\frac{1}{x^2+yz +1})=3-\sum\frac{1}{x^2+yz +1}$ $=>$ $\frac{x^2+yz}{x^2+yz +1}+\frac{y^2+zx}{y^2+zx+1}+\frac{z^2+xy}{z^2+xy+1}=3-\sum\frac{1}{x^2+yz +1}$ $...(1)$ $\sum\frac{1}{x^2+yz +1}=\frac{1}{x^2+yz +1}+\frac{1}{y^2+zx +1}+\frac{1}{z^2+xy +1}$ By $Cauchy$ $Schwarz$ $Inequality$ we have: $\frac{1}{x^2+yz +1}+\frac{1}{y^2+zx +1}+\frac{1}{z^2+xy +1}\ge \frac{(1+1+1)^2}{x^2+yz +1+y^2+zx +1+z^2+xy +1}$ $=>$ $\sum\frac{1}{x^2+yz +1}\ge \frac{3^2}{x^2+y^2+z^2+xy+yz+xz+1+1+1}$ $=>$ $\sum\frac{1}{x^2+yz +1}\ge \frac{9}{x^2+y^2+z^2+xy+yz+xz+3}$ $...(2)$ From $AM-GM$ we get: $x^2+y^2\ge2\sqrt{x^2*y^2}=2\sqrt{(xy)^2)}=2xy$ $y^2+z^2\ge2\sqrt{y^2*z^2}=2\sqrt{(yz)^2)}=2yz$ $x^2+z^2\ge2\sqrt{x^2*z^2}=2\sqrt{(xz)^2)}=2xz$ $$=>$$$x^2+y^2+y^2+z^2+x^2+z^2\ge2xy+2yz+2xz$ $2x^2+2y^2+2z^2\ge2xy+2yz+2xz$ $x^2+y^2+z^2\ge xy+yz+xz$ $=>$ $\frac{1}{x^2+y^2+z^2}\leq\frac{1}{xy+yz+xz}$ $=>$ $\frac{1}{xy+yz+xz}\ge\frac{1}{x^2+y^2+z^2}$ $...(3)$ Combining $(2)$ with $(3)$ we get: $\sum\frac{1}{x^2+yz +1}\ge \frac{9}{x^2+y^2+z^2+xy+yz+xz+3}\ge\frac{9}{x^2+y^2+z^2+x^2+y^2+z^2+3}=\frac{9}{2x^2+2y^2+2z^2++3}$$=$ $=$$\frac{9}{2(x^2+y^2+z^2)+3}=\frac{9}{2*3+3}==\frac{9}{6+3}=\frac{9}{9}=1$ $=>$ $\sum\frac{1}{x^2+yz +1}\ge1$ $...(4)$ Now combining $(1)$ with $(4)$ we get: $\frac{x^2+yz}{x^2+yz +1}+\frac{y^2+zx}{y^2+zx+1}+\frac{z^2+xy}{z^2+xy+1}=3-\sum\frac{1}{x^2+yz +1}\ge3-1=2$ $=>$ $\frac{x^2+yz}{x^2+yz +1}+\frac{y^2+zx}{y^2+zx+1}+\frac{z^2+xy}{z^2+xy+1}\ge2$

Let $f(x)=\frac{x}{x+1}\Longrightarrow f'(x)=\frac{1}{(x+1)^2}\Longrightarrow f''(x)=-\frac{2}{(x+1)^3}<0, \forall x \in \mathbb{R}^+$ thus $f$ is concave. Furthermore $\sum_{cyc}\frac{x^2+yz}{x^2+yz+1}=\sum_{cyc}f(t_1)\le3f\left(\frac{\sum_{cyc}t_1}{3}\right)$ where $t_1=x^2+yz, t_2=y^2+zx\text{ and } t_3=z^2+xy$. Thus the inequality is equivalent to $3f\left(\frac{3+\sum_{cyc}xy}{3}\right)=\frac{9+3\sum_{cyc}xy}{6+\sum_{cyc}xy}\le2\Longleftrightarrow 12+2\sum_{cyc}xy\ge9+3\sum_{cyc}xy$ $\therefore 3\ge\sum_{cyc}xy\Longleftrightarrow \sum_{cyc}x^2\ge\sum_{cyc}xy$ $\blacksquare$.

The inequality can be transformed as follows: $$\sum\frac{x^2+yz}{x^+yz+1}=\sum\frac{x^2+yz+1-1}{x^+yz+1}=3-(\sum \frac{1}{x^2+yz+1})\leq2$$$$\Rightarrow \sum\frac{1}{x^2+yz+1}\geq1$$By Titu's lemma we have: $$\sum\frac{1}{x^2+yz+1}\geq \frac{(1+1+1)^2}{\sum x^2+yz+1}=\frac{9}{x^2+yz+1+y^2+xz+1+z^2+xy+1}$$It suffices to prove that: $\frac{9}{x^2+y^2+z^2+xy+yz+xz+3}\geq 1$ $\Rightarrow$ $x^2+y^2+z^2+xy+yz+xz+3\leq9$ It's quite easy to prove that $xy+yz+xz\leq x^2+y^2+z^2=3$ (1) from which we have: $$x^2+y^2+z^2+xy+yz+xz+3\leq3+3+3=9$$