It suffice to show that $\rm \sum a\sqrt{b^2+c^2+bc}\geq \sqrt{3}(ab+bc+ac)$ $\rm LHS=\sum \sqrt{a^2b^2+b^2c^2+ab\cdot ac}$ We set $\rm ab=x,ac=y,bc=z$.Now we must show that $\rm \sum \sqrt{x^2+y^2+xy}\geq \sqrt{3}(x+y+z)$. We have that $\rm \sqrt{x^2+y^2+xy}\geq \dfrac{\sqrt{3}}{2}(x+y)\Leftrightarrow 4(x^2+y^2+xy)\geq 3(x^2+y^2+xy)\Leftrightarrow x^2+y^2\geq 2xy$ which is true, so $\rm LHS\geq \sum \dfrac{\sqrt{3}}{2}(x+y)=\sqrt{3}(x+y+z)$ QED

parmenides51 wrote: Let $a, b,c$ be positive real numbers such that $ab + bc + ca = 1$. Prove that $$a\sqrt{b^2 + c^2 + bc} + b\sqrt{c^2 + a^2 + ca} + c\sqrt{a^2 + b^2 + ab} \ge \sqrt3$$ https://artofproblemsolving.com/community/c6h82033p3582507 arqady wrote: sqing wrote: 2..Let $a,b,c$ be positive real numbers such that $bc+ca+ab=1$ . Prove that $$a\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}+\sqrt{a^2+ab+b^2}\ge \sqrt{3}.$$ $$\sum_{cyc}a\sqrt{b^2+bc+c^2}\geq\sum_{cyc}\frac{\sqrt3a(b+c)}{2}=\sqrt3$$

Let $a, b,c$ be real numbers . Prove that $$\sqrt{a^2-ab+b^2-3a+3b+3} +\sqrt{b^2-bc+c^2-3b+3c+3} +\sqrt{c^2-ca+a^2-3c+3a+3} \ge 3\sqrt3$$ ZDSX,9(2020) Let $x, y, z>0.$Prove that $$3(x+y+z)<\sum_{ cyc }\sqrt{x^{2}+4y^{2}+z^{2}+4xy+4yz-2xz}<6(x+y+z)$$

i have harder version of the same problem

q1

solved also here (as p3)

Let $a, b,c$ be positive real numbers such that $ab + bc + ca = 1$. Prove that $$a\sqrt{b^2 + c^2-bc} + b\sqrt{c^2 + a^2-ca} + c\sqrt{a^2 + b^2-ab} \ge 1$$ $$\sqrt{a(b^2 + c^2 + bc)} +\sqrt{b(c^2 + a^2 + ca)} +\sqrt{c(a^2 + b^2 + ab)} \ge \sqrt[4]{27}$$ $$\sqrt{a(b^2 + c^2 -bc)} +\sqrt{b(c^2 + a^2-ca)} +\sqrt{c(a^2 + b^2-ab)} \ge \sqrt[4]{3}$$

sqing wrote: Let $a, b,c$ be positive real numbers such that $ab + bc + ca = 1$. Prove that $$a\sqrt{b^2 + c^2-bc} + b\sqrt{c^2 + a^2-ca} + c\sqrt{a^2 + b^2-ab} \ge 1$$ Note that $$b^2+c^2-bc\ge \frac{(b+c)^2}{4}\iff \boxed{3(b-c)^2\ge 0.}$$ Therefore, $$\boxed{\sum_{cyc}a\sqrt{b^2 + c^2-bc} \ge bc+ca+ab=1.}$$ Equality holds iff $a=b=c=\frac{1}{\sqrt 3}.$ $\textcolor{green}{\textit{QED.}}$