Problem: If {a, b, c} are positive numbers, then show that
\displaystyle{{\frac{b^2 + c^2}{b + c}} + {\frac{c^2 + a^2}{c + a}} + {\frac{a^2 + b^2}{a + b}} \ge {a + b + c}}.

Solution: We know that {a, b, c > 0}

L.H.S = {\displaystyle{{\frac{b^2 + c^2}{b + c}} + {\frac{c^2 + a^2}{c + a}} + {\frac{a^2 + b^2}{a + b}}}}

{\ge} {\displaystyle{\frac{{\frac{b^2 + c^2 + 2bc}{b + c}} + {\frac{c^2 + a^2 + 2ac}{c + a}} + {\frac{a^2 + b^2 + 2ab}{a + b}}}{2}}} [ as {b + c}, {c + a}, {a + b > 0} & {x^2 + y^2 > 2xy} for all {x + y {\in} |R} ]

= {a + b + c}
= R.H.S [ proved ]