parmenides51 wrote: Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. Prove that $$2 \left( \frac{ab}{a + b} +\frac{bc}{b + c} +\frac{ca}{c+ a}\right)+ 1 \ge 6(ab + bc + ca)$$ Trần Nam Dũng https://artofproblemsolving.com/community/c6h1460136p16699428 $$\displaystyle \because \sum \frac{ab(a-b)^2}{(b+c)(a+c)} \geqslant 0$$