The problem suffices to find the largest constant $k$ such that for $a,b,c \ge 0:a+b+c=3$ we always have ${a^2} + {b^2} + {c^2} + kabc \ge k + 3$. $(*)$ Let $b=c$, we have \[\begin{array}{l} a + 2b = 3 \\ \frac{{{a^2} + 2{b^2} - 3}}{{1 - a{b^2}}} \ge k \\ \end{array}\] which means \[\frac{{{{\left( {3 - 2b} \right)}^2} + 2{b^2} - 3}}{{1 - \left( {3 - 2b} \right){b^2}}} \ge k,\forall b \in \left[ {0,\frac{3}{2}} \right]\] And $k \le \frac{3}{2}$ can be infered. We only need to prove $(*)$ in the case $k = \frac{3}{2}$, which is equivalent to \[\begin{array}{l} 2\left( {{a^2} + {b^2} + {c^2}} \right) + 3abc \ge 9 \\ \Leftrightarrow 2\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2}} \right) + 9abc \ge {\left( {a + b + c} \right)^3} \\ \end{array}\] Expanding the last inequality and we get Schur in degree 3, which is obviously true. Done.

nguyenhung wrote: The problem suffices to find the largest constant $k$ such that for $a,b,c \ge 0:a+b+c=3$ we always have ${a^2} + {b^2} + {c^2} + kabc \ge k + 3$. $(*)$ Let $b=c$, we have \[\begin{array}{l} a + 2b = 3 \\ \frac{{{a^2} + 2{b^2} - 3}}{{1 - a{b^2}}} \ge k \\ \end{array}\] which means \[\frac{{{{\left( {3 - 2b} \right)}^2} + 2{b^2} - 3}}{{1 - \left( {3 - 2b} \right){b^2}}} \ge k,\forall b \in \left[ {0,\frac{3}{2}} \right]\] And $k \le \frac{3}{2}$ can be infered. We only need to prove $(*)$ in the case $k = \frac{3}{2}$ Why??

opa wrote: Why?? We call a $k$ is "VIP" if it has the following properties: "for all nonnegative real numbers $a,b,c: \; a^2+b^2+c^2+kabc=k+3$, we always have $a+b+c \le 3$." We also call a $k$ is "interesting" if it has the following properties: "for all nonnegative real numbers $a,b,c: \; a+b+c=3$, we always have $a^2+b^2+c^2+kabc \ge k+3$." We will demonstrate that a $k$ is "VIP" if and only if it is "interesting". For every VIP $k$, if we change $c$ to $c'$ such that $a+b+c'=3$ then $a^2+b^2+c'^2+kabc' \ge a^2+b^2+c^2+kabc \ge k+3$. Hence, that $k$ is also $interesting$. For every interesting $k$, if we change $c$ to $c'$ such that $a^2+b^2+c'^2+kabc'=k+3$ then we have $a+b+c' \le 3$, which means $k$ is also "VIP". Done.