$a+b\ge c \implies ac+bc\ge c^2 \implies 2(ab+bc+ca)\ge a^2 + b^2 + c^2 \implies (a+b)\sqrt{ab}+(a+c)\sqrt{ac}+(b+c)\sqrt{bc} \geq 2(ab+bc+ca) \ge (a+b+c)^2/2$

Thanks. The International School Olympiad “Caucasus Mathematical Olympiad” was founded in 2015 by the initiative of Daud Mamiy, Dean of the Faculty of Mathematics and Computer Sciences at Adyghe State University. The organizers of the Olympiad are Ministry of Education and Science of the Republic of Adygea, Adyghe State University and School of Mathematics and Natural Science. http://cmo.adygmath.ru/en/node/52

bigant146 wrote: Let $a, b, c$ be the lengths of sides of a triangle. Prove the inequality $$(a+b)\sqrt{ab}+(a+c)\sqrt{ac}+(b+c)\sqrt{bc} \geq (a+b+c)^2/2.$$ Let $a, b, c$ be the lengths of sides of a triangle. Then $$ \frac{(a+b+c)^2}{2}<(a+b)\sqrt{ab}+(a+c)\sqrt{ac}+(b+c)\sqrt{bc}< \frac{(a+b+c)^2}{2}+ \frac{abc}{2}.$$

bigant146 wrote: Let $a, b, c$ be the lengths of sides of a triangle. Prove the inequality $$(a+b)\sqrt{ab}+(a+c)\sqrt{ac}+(b+c)\sqrt{bc} \geq (a+b+c)^2/2.$$ Let $a, b, c$ be the lengths of sides of a triangle. Then $$a\sqrt{ab}+b\sqrt{bc}+c\sqrt{ca} \geq (a+b+c)^2/4.$$

Grotex wrote: Let $a, b, c$ be the lengths of sides of a triangle. Then $$a\sqrt{ab}+b\sqrt{bc}+c\sqrt{ca} \geq (a+b+c)^2/4.$$ Ravi substitution, squaring and C-S.

bigant146 wrote: Let $a, b, c$ be the lengths of sides of a triangle. Prove the inequality $$(a+b)\sqrt{ab}+(a+c)\sqrt{ac}+(b+c)\sqrt{bc} \geq (a+b+c)^2/2.$$ AM-GM, then Ravi Substitution and we're done

Constant $\frac{1}{2}$ on the right is unimprovable. Take $x,y\to 0^{+}$

Let $a, b, c$ be the lengths of sides of a triangle. Prove the inequality $$(a+b)\sqrt{ab}+(a+c)\sqrt{ac}+(b+c)\sqrt{bc} \geq 2(ab+bc+ca)> \frac{(a+b+c)^2}{2}.$$