Inequality of squares (Tomato subjective 80)

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 ]