Problem 6. Let $a, b, c$ be nonnegative real numbers. Find the maximum of $k$ such that the inequality
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+k \cdot \frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \geq k+\frac{3}{2}$$
Solution:
We have that equality $\sum \frac{a}{b+c}-\frac{3}{2}=\frac{1}{2} \sum \frac{(a-b)^{2}}{(a+c)(b+c)}$.
That inequality can be rewritten as
$$\sum \frac{(a-b)^{2}}{(a+c)(b+c)} \geq k \sum \frac{(a-b)^{2}}{a^{2}+b^{2}+c^{2}} \Leftrightarrow \sum(a-b)^{2}\left(\frac{a^{2}+b^{2}+c^{2}}{(a+c)(b+c)}-k\right) \geq 0$$

Let $b=c$ then $k$ needs to satisfy the condition for all $a, b \geq 0$
$$k \leq \frac{a^{2}+b^{2}+c^{2}}{(a+c)(b+c)}=\frac{a^{2}+2 b^{2}}{2 b(a+b)}$$

We have that $\frac{a^{2}+2 b^{2}}{2 b(a+b)} \geq \frac{\sqrt{3}-1}{2}$.
We will prove that $k=\frac{\sqrt{3}-1}{2}$ is the best constant. WLOG, assume $a \geq b \geq c$ so
$$S_{a}=\frac{a^{2}+b^{2}+c^{2}}{(a+c)(a+b)}-k, S_{b}=\frac{a^{2}+b^{2}+c^{2}}{(b+a)(b+c)}-k, S_{c}=\frac{a^{2}+b^{2}+c^{2}}{((c+a)(c+b)}-k .$$

We easily see that $S_{c} \geq S_{b} \geq S_{a}$, also
$$S_{b}+S_{a}=\frac{\left(a^{2}+b^{2}+c^{2}\right)(a+b+2 c)}{(a+b)(b+c)(c+a)}-2 k$$

Let $t=\frac{a+b}{2}$ we have
$$S_{a}+S_{a} \geq \frac{\left(2 t^{2}+c^{2}\right)(2 t+2 c)}{2 t(t+c)^{2}}-2 k=\frac{2 t^{2}+c^{2}}{t(t+c)}-2 k$$

The equality holds when $a=b=c$ or $a=b=\frac{\sqrt{3}+1}{2} c$ and any cyclic permutation, and $k=\frac{\sqrt{3}-1}{2}$.