Example 1.11.5. Let $a, b, c$ be non-negative real numbers. For each $k, l \geq 0$, find the maximal and minimal value of the expression
$$\frac{a}{k a+l b+c}+\frac{b}{k b+l c+a}+\frac{c}{k c+l a+b} \text {. }$$
SOLUTION. First we will examine the expression in the case $b=1, c=0$. Denote
$$f(a)=\frac{a}{k a+l}+\frac{1}{k+a}$$
then we get
$$f^{\prime}(a)=\frac{l}{(k a+l)^{2}}-\frac{1}{(k+a)^{2}}$$

The equation $f^{\prime}(a)=0$ has exactly one positive real root $a=\sqrt{l}$. So, if $l>k^{2}$ then
$$\begin{aligned}
\min f(a) & =f(\sqrt{l})=\frac{2}{k+\sqrt{l}} \\
\sup f(a) & =\lim _{a \rightarrow+\infty} f(a)=\frac{1}{k}
\end{aligned}$$

Otherwise, if $l<k^{2}$ then
$$\begin{array}{c}
\min f(a)=f(0)=\frac{1}{k} \\
\max f(a)=f(\sqrt{l})=\frac{2}{k+\sqrt{l}}
\end{array}$$

For $l=k^{2}$ then we have
$$f(a)=\frac{a}{k a+k^{2}}+\frac{1}{k+a}=\frac{1}{k} .$$

Denote by
$$F=\frac{a}{k a+l b+c}+\frac{b}{k b+l c+a}+\frac{c}{k c+l a+b} .$$

According to the previous results, we have the conclusion
$$\begin{array}{l}
\min F=\min \left\{\frac{3}{k+l+1} ; \frac{1}{k} ; \frac{2}{k+\sqrt{l}}\right\} \\
\sup F=\max \left\{\frac{3}{k+l+1} ; \frac{1}{k} ; \frac{2}{k+\sqrt{l}}\right\}
\end{array}$$