From the condition we have $x+y+z \ge 3$. We proceed with Cauchy: $(x^2+y+z)(1+y+z) \ge (x+y+z)^2$ from where we get the left side of the starting inequality is $ \le \frac{2(x+y+z)+3}{(x+y+z)^2} \le 1$ which is equivalent to proving $(t-3)(t+1)\ge 0 $ where $t=x+y+z \ge 3$ so we are done ( the condition is equivalent to $x^2+y^2+z^2 \le xyz(x+y+z)$ $\frac{(x+y+z)^2}{3} \le x^2+y^2+z^2 \le xyz(x+y+z) \le \frac{(x+y+z)^4}{3 \cdot 9}$ now from the first and last inequality we get $t \ge 3$)

$\color{red}{\textit{\textbf{Proof:}}}$ Since by Cauchy-Schwarz Inequality, \[(x+y+z)(3xyz)\ge \left(\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}\right)(xyz+xyz+xyz)\ge (x+y+z)^2,\]we have $3xyz\ge x+y+z \ge 3\sqrt[3] {xyz}$ by AM-GM, which gives $xyz\ge 1$ and $x+y+z\ge 3\sqrt[3] {xyz}\ge 3$. By Cauchy-Schwarz Inequality again, \[\sum_{cyc} \frac{1}{x^2+y+z}\le \sum_{cyc} \frac{1+y+z}{(x+y+z)^2}=\frac{2(z+y+z)+3}{(x+y+z)^2} \le 1 \iff (x+y+z-3)(x+y+z+1)\ge 0\]which is true. $\quad \blacksquare$

$\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}\le x+y+z \implies (x+y+z)(xyz + xyz + xyz) \ge (\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy})(xyz+xyz+xyz) \le (x+y+z)^2 \implies xyz \ge 1 \implies x+y+z \ge 3$. $\frac{1}{x^2+y+z}+\frac{1}{y^2+z+x}+\frac{1}{z^2+x+y} \le \frac{3+2(x+y+z)}{(x+y+z)^2}$ so we need to prove $(x+y+z)^2 - 2(x+y+z) - 3 \ge 0$ or $(x+y+z)^2 \ge 3(x+y+z)$ which is true.

Let $x,y,z$ be positive real numbers with $x+y+z \ge 3$. Prove that$$\frac{1}{x^2+y+z}+\frac{1}{y^2+z+x}+\frac{1}{z^2+x+y} \le 1$$46th Austrian Mathematical Olympiad National Competition

Same as other solutions, posting for storage The condition is basically $x+y+z \ge 3$. By $C-S$ $(x^2+y+z)(1+y+z) \ge (x+y+z)^2$ hence $$\sum \frac{1}{x^2+y+z}= \sum \frac{1+y+z}{(x^2+y+z)(1+y+z)} \le \sum \frac{1+y+z}{(x+y+z)^2}= \frac{2(x+y+z)+3}{(x+y+z)^2} \leftrightarrow$$$$2(x+y+z)+3 \le 2(x+y+z)+(x+y+z)\le (x+y+z)^2$$dividing by $(x+y+z)$ gives $3 \le x+y+z$ which is true $\blacksquare$

The proof of the ineq equal to Prove:$\sum{\frac{x}{x+2}} \ge 1(?)$ By titu (x+y+z)^2>=2(x+y+z)+x^2+y^2+z^2 And we are done

$xyz\geq \frac{\sum{x^2}}{\sum{x}}\geq \frac{\sum{x}}{3}\implies (\sum{x})^2\leq 9(xyz)^2$ \[\sum{\frac{xyz}{x^2+y+z}}\leq\sum{\frac{xyz}{3\sqrt[3]{x^2yz}}}=\frac{\sum{\sqrt[3]{xy^2z^2}}}{3}\leq \frac{\sqrt[3]{\sum{x^2yz}.\sum{x}.3}}{3}\leq \frac{\sqrt[3]{xyz.(\sum{x})^2}}{\sqrt[3]{9}}\leq \frac{\sqrt[3]{9(xyz)^3}}{\sqrt[3]{9}}=xyz\]