2.2. Determine the smallest constant $k>0$ such that
$$
\frac{a b}{a+b+2 c}+\frac{b c}{b+c+2 a}+\frac{c a}{c+a+2 b} \leq k(a+b+c) \text {, }
$$
for all $a, b, c>0$.
2.2 The smallest such value of $k$ is $1 / 4$. First note that for $x, y>0$,
$$
\frac{1}{x+y}=\frac{4 x y}{x+y} \cdot \frac{1}{4 x y} \leq \frac{(x+y)^{2}}{(x+y)} \cdot \frac{1}{4 x y}=\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right),
$$
with equality if and only if $x=y$. We then have $\frac{a b}{a+b+2 c}+\frac{b c}{b+c+2 a}+\frac{c a}{c+a+2 b}$
$$
\leq \frac{a b}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)+\frac{b c}{4}\left(\frac{1}{c+a}+\frac{1}{b+a}\right)+\frac{c a}{4}\left(\frac{1}{c+b}+\frac{1}{a+b}\right)=\frac{1}{4}(a+b+c),
$$
with equality if and only if $a+b=b+c=c+a$, that is $a=b=c$.