SELECTION EXAMINATION

Since $a+b+c=1$, we have $a+1=2a+b+c$ and $1-a=b+c$. Hence we have the equivalent inequality $$(2a+b+c)\sqrt{2a(b+c)}+(2b+c+a)\sqrt{2b(c+a)}+(2c+a+b)\sqrt{2c(a+b)}\ge 8(ab+bc+ca)$$ From the inequality of arithmetic and geometric mean we get: $2a+b+c\ge 2\sqrt{2a(b+c)}$ and therefore $$(2a+b+c)\sqrt{2a(b+c)}\ge 2(2a(b+c))(1)$$ Similarly we have $$(2b+c+a)\sqrt{2b(c+a)}\ge 2(2b(c+a))(2),$$ $$(2c+a+b)\sqrt{2c(a+b)}\ge 2(2c(a+b))(3).$$ Summing by parts (1), (2) and (3) we get the wanted inequality. The equality holds if and only if $$2a=b+c,2b=c+a,2c=a+b\iff a=b=c=\frac{1}{3}.$$