Example 1.10.6. Let \(a, b, c\) be non-negative real numbers such that \(a+b+c=3\). For each \(k \geq 0\), find the maximum and minimum value of
\[
a^{2}(k b-c)+b^{2}(k c-a)+c^{2}(k a-b) .
\]
SOLUTION. As in the preceding solutions, we will first consider the case $c=0$. Denote
$$k a^{2} b-a b^{2}=k a^{2}(3-a)-a(3-a)^{2}=-(k+1) a^{3}+3(k+2) a^{2}-9 a=f(a)$$
then we get
$$f^{\prime}(a)=-3(k+1) a^{2}+6(k+2) a-9$$

The equation $f^{\prime}(a)=0$ has exactly two positive real roots (in $[0,3]$ )
$$a_{1}=\frac{k+2-\sqrt{k^{2}+k+1}}{k+1} ; a_{2}=\frac{k+2+\sqrt{k^{2}+k+1}}{k+1} .$$

Therefore, we infer that
$$\begin{array}{l}
\min _{a \in[0,3]} f(a)=f\left(a_{1}\right)=\frac{2\left(k^{2}+k+1\right)}{(k+1)^{2}}\left(k+2-\sqrt{k^{2}+k+1}\right)-\frac{3(k+2)}{k+1}=m \\
\max _{a \in[0,3]} f(a)=f\left(a_{1}\right)=\frac{2\left(k^{2}+k+1\right)}{(k+1)^{2}}\left(k+2+\sqrt{k^{2}+k+1}\right)-\frac{3(k+2)}{k+1}=M
\end{array}$$

The inequality is cyclic, and therefore, due to CD3 theorem, we conclude
$$\begin{array}{l}
\min a^{2}(k b-c)+b^{2}(k c-a)+c^{2}(k a-b)=\min \{3(k-1) ; m\} \\
\max a^{2}(k b-c)+b^{2}(k c-a)+c^{2}(k a-b)=\max \{3(k-1) ; M\}
\end{array}$$