Example 1.11.2. Let $a, b$, c be non-negative real numbers. For each $k \geq 0$, find the minimum of the expression
$$\frac{a+k b}{c+k b}+\frac{b+k c}{a+k c}+\frac{c+k a}{b+k a}$$
SOLUTION. For $b=1, c=0$, the expression becomes
$$f(a)=\frac{a+k}{k}+\frac{1}{a}+\frac{k a}{1+k a}=2+\frac{a}{k}+\frac{1}{a}-\frac{1}{1+k a}$$

If $k \leq 1$ then, according to AM-GM inequality, we have
$$f(a) \geq 2+\frac{2}{\sqrt{k}}-1 \geq 3$$

We will now consider the main case, when $k \geq 1$. Clearly,
$$f(a)=2+\frac{a}{k}+\frac{1}{a}-\frac{1}{1+k a} \geq 2+\frac{a}{k}>2$$

However, 2 is not the minimal value of $f(a)$. To find this value, we have to use derivatives. We have first
$$f'(a)=\frac{1}{k}-\frac{1}{a^{2}}+\frac{k}{(1+k a)^{2}}$$

The equation $f'(a)=0$ is equivalent to
$$k^{2} a^{4}+2 k a^{3}-\left(k^{3}+k^{2}-1\right) a^{2}-2 k^{2} a-k=0 .$$

It is easy to check that the equation has exactly one real root $a_{0}$, so
$$\min f(a)=f\left(a_{0}\right)=2+\frac{a_{0}}{k}+\frac{1}{a_{0}}-\frac{1}{1+k a_{0}}$$

According to the CD3 theorem (for fractions), we conclude that
$$\min \left\{\frac{a+k b}{c+k b}+\frac{b+k c}{a+k c}+\frac{c+k a}{b+k a}\right\}=\min \left\{3 ; 2+\frac{a_{0}}{k}+\frac{1}{a_{0}}-\frac{1}{1+k a_{0}}\right\}$$