# 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 ]