Selected Problems from the Final Round of National Olympiad

W.l.o.g., assume $a\ge b\ge c$. Then $$\begin{array}{rl}\frac{a^2+bc}{b+c}+\frac{b^2+ca}{c+a}+\frac{c^2+ab}{a+b}& =\\ & =\frac{a^2+(b+c)c-c^2}{b+c}+\frac{b^2+(c+a)a-a^2}{c+a}+\frac{c^2+(a+b)b-b^2}{a+b}=\\ & =\frac{a^2-c^2}{b+c}+c+\frac{b^2-a^2}{c+a}+a+\frac{c^2-b^2}{a+b}+b\ge a+b+c,\end{array}$$ since $$\begin{array}{rl}\frac{a^2-c^2}{b+c}+\frac{b^2-a^2}{c+a}+\frac{c^2-b^2}{a+b}& =\frac{a^2-b^2+b^2-c^2}{b+c}+\frac{b^2-a^2}{c+a}+\frac{c^2-b^2}{a+b}=\\ & =(a^2-b^2)\left(\frac{1}{b+c}-\frac{1}{c+a}\right)+(b^2-c^2)\left(\frac{1}{b+c}-\frac{1}{a+b}\right)=\\ & =\frac{(a^2-b^2)(a-b)}{(b+c)(c+a)}+\frac{(b^2-c^2)(a-c)}{(b+c)(a+b)}\ge 0.\end{array}$$

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Alternative solution.

Rearranging and transforming the expression gives $$\begin{array}{rl}\frac{a^2+bc}{b+c}-a+\frac{b^2+ca}{c+a}-b+\frac{c^2+ab}{a+b}-c& =\\ & =\frac{a^2+bc-ab-ac}{b+c}+\frac{b^2+ca-bc-ba}{c+a}+\frac{c^2+ab-ca-cb}{a+b}=\\ & =\frac{(a-b)(a-c)}{b+c}+\frac{(b-c)(b-a)}{c+a}+\frac{(c-a)(c-b)}{a+b}=\\ & =\frac{(a^2-b^2)(a^2-c^2)+(b^2-c^2)(b^2-a^2)+(c^2-a^2)(c^2-b^2)}{(b+c)(c+a)(a+b)}=\\ & =\frac{a^4+b^4+c^4-a^2{b}^2-a^2{c}^2-b^2{c}^2}{(b+c)(c+a)(a+b)}=\\ & =\frac{(a^2-b^2{)}^2+(c^2-a^2{)}^2+(b^2-c^2{)}^2}{2(b+c)(c+a)(a+b)}\ge 0.\end{array}$$