## Task 1.

Determine the smallest real number $D$ such that the inequality

$$
\frac{a+b}{a+2 b}+\frac{b+c}{b+2 c}+\frac{c+a}{c+2 a}<D
$$

holds for all positive real numbers $a, b$, and $c$.
## Solution.

The first addend in the inequality can be written as follows:

$$
\frac{a+b}{a+2 b}=\frac{a+2 b}{a+2 b}-\frac{b}{a+2 b}=1-\frac{b}{a+2 b}=1-\frac{1}{\frac{a}{b}+2}=1-\frac{1}{x+2}
$$

where we denote $x=\frac{a}{b}$. If we denote $y=\frac{b}{c}$ and $z=\frac{c}{a}$, the desired inequality is equivalent to the inequality

$$
\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}>3-D
$$

under the condition $x y z=1$.

Let's denote the expression on the left side as $X$ and choose an arbitrary natural number $n$. If we substitute $x=n, y=n, z=\frac{1}{n^{2}}$, we get:

$$
X=\frac{1}{2+\frac{1}{n^{2}}}+\frac{2}{n+2}=\frac{n^{3}+6 n^{2}+2}{\left(2 n^{2}+1\right)(n+2)}=\frac{1}{2}+\frac{8 n^{2}-n+2}{2\left(2 n^{3}+4 n^{2}+n+2\right)}
$$

The number $\frac{8 n^{2}-n+2}{2\left(2 n^{3}+4 n^{2}+n+2\right)}$ can be an arbitrarily small positive number, so $X$ can be arbitrarily close to $\frac{1}{2}$.

It remains to prove the inequality: $\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}>\frac{1}{2}$.

By multiplying by the common denominator, we see that this inequality is equivalent to

$$
2(y+2)(z+2)+2(z+2)(x+2)+2(x+2)(y+2)>(x+2)(y+2)(z+2)
$$

or

$$
2(x y+y z+z x)+4(x+y+z)+24>x y z+2(x y+y z+z x)+4(x+y+z)+8
$$

i.e., $16>x y z=1$, which is obviously true.

Therefore, $3-D=\frac{1}{2}$, so $D=\frac{5}{2}$.