For all positive real numbers a, b, and c, prove that
\frac{(a^2 + bc)^2}{b + c} + \frac{(b^2 + ca)^2}{c + a} + \frac{(c^2 + ab)^2}{a + b} \ge \frac{2abc(a+b+c)^2}{ab + bc + ca}.
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Doing AM-GM
LHS \ge \sum \frac{ (2 \sqrt {a^2bc})^2 }{b+c} \Rightarrow LHS \ge \sum \frac{4a^2bc }{b+c}
Divide by 2abc it’s suffices to prove \sum \frac{2a}{b+c} \ge \frac{(a+b+c)^2}{ab+bc+ca}
Now doing Cauchy–Schwarz
\sum \frac {2a}{b+c} \ge 2 \frac{(a+b+c)^2}{2(ab+bc+ca)} = \frac{(a+b+c)^2}{(ab+bc+ca)}
Done.