62nd Ukrainian National Mathematical Olympiad

If we assume that $b+c=0$, then we have $$ab+bc+ac=a(b+c)+ac=ac=0,$$ which contradicts the condition. Thus $(a+b)(b+c)(c+a)\ne 0$. Since $$(a+b+c)(a+b)=a^2+ab+b^2+ab+bc+ac=a^2+ab+b^2,$$ we have $\frac{1}{a+b}=(a+b+c)\cdot \frac{1}{a^2+ab+b^2}$. Then, adding three similar equations, we get $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}=(a+b+c)\left(\frac{1}{a^2+ab+b^2}+\frac{1}{b^2+bc+c^2}+\frac{1}{a^2+ac+c^2}\right).$$ It is easy to see that each of the expressions of the form $a^2+ab+b^2=\frac{a^2+(a+b{)}^2+b^2}{2}>0$, which implies the desired statement.

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Alternative solution.

Similarly to the previous solution, we verify that the given expressions are defined correctly, that is, $(a+b)(b+c)(c+a)\ne 0$. From the given equation, we write $c=-\frac{ab}{a+b}$ and substitute this expression into the product: $$\begin{gathered} (a+b+c)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)=(a+b-\frac{ab}{a+b})\left(\frac{1}{a+b}+\frac{1}{b-\frac{ab}{a+b}}+\frac{1}{a-\frac{ab}{a+b}}\right)= \\ =\left(\frac{(a+b{)}^2-ab}{a+b}\right)\left(\frac{1}{a+b}+\frac{a+b}{b^2}+\frac{a+b}{a^2}\right)= \\ =\frac{a^2+ab+b^2}{a+b}\cdot \frac{a^2{b}^2+(a+b{)}^2{a}^2+(a+b{)}^2{b}^2}{(a+b)a^2{b}^2}= \\ =(a^2+ab+b^2)\cdot \frac{a^2{b}^2+(a+b{)}^2{a}^2+(a+b{)}^2{b}^2}{(a+b{)}^2{a}^2{b}^2}>0, \end{gathered}$$ since $a^2+ab+b^2=\frac{a^2+(a+b{)}^2+b^2}{2}>0$, and the second factor is also obviously positive. If the product of two expressions is always positive, then they have the same sign, which is what we needed to prove.