Let $\triangle ABC$ have sides $BC=a$, $CA=b$, and $AB=c$. Let $m_a$, $m_b$, $m_c$ be the medians to $BC$, $CA$, and $AB$, respectively. Prove that $$(ab+bc+ca)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 2\sqrt{3}(m_a+m_b+m_c)$$
My trying: $$\Leftrightarrow (ab+bc+ca)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 3\sqrt{3}\sqrt{a^2+b^2+c^2}$$ Because: $$m_a+m_b+m_c\leq \sqrt{3(m_{a}^{2}+m_{b}^{2}+m_{c}^{2})}=\frac{3}{2}\sqrt{a^2+b^2+c^2}$$ $$\Leftrightarrow \left(p^2+4Rr+r^2\right)\left(\frac{4R+r}{pr} \right)\geq 3\sqrt{3}\sqrt{2p^2-8Rr-2r^2}$$ I square it, but that doesn't help.
The hint:
Draw the $\Delta ABC$ with medians $m_b=BD$ and $CE=m_a$.
Thus, by the Ptolemy's theorem for the equilateral $BCDE$ we obtain: $$BE\cdot DC+BC\cdot ED\geq BD\cdot CE$$ or $$\frac{bc}{4}+\frac{a^2}{2}\geq m_am_b.$$ Hence, $$\sum_{cyc}m_a=\sqrt{\sum_{cyc}\left(m_a^2+2m_bm_c\right)}\leq\sqrt{\sum_{cyc}\left(m_a^2+2\left(\frac{bc}{4}+\frac{a^2}{2}\right)\right)}=$$ $$=\sqrt{\sum_{cyc}\left(\frac{1}{4}(2b^2+2c^2-a^2)+2\left(\frac{bc}{4}+\frac{a^2}{2}\right)\right)}=\frac{1}{2}\sqrt{\sum_{cyc}(7a^2+2ab)}.$$ It est, it's enough to prove that $$\frac{(ab+ac+bc)^4}{3a^2b^2c^2}\geq\sum_{cyc}(7a^2+2ab).$$ Now, let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x$, $y$ and $z$ are positives and we need to prove that $$\left(\sum_{cyc}(x^2+3xy)\right)^4\geq3\prod_{cyc}(x+y)^2\sum_{cyc}(14x^2+14y^2+2x^2+6xy)$$ or $$\left(\sum_{cyc}(x^2+3xy)\right)^4\geq12\prod_{cyc}(x+y)^2\sum_{cyc}(4x^2+5xy)$$ or $$\sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)+$$ $$+24xyz\sum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)\geq0.$$ Now, $$\sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)=$$ $$=\frac{1}{2}\sum_{cyc}(x^8+y^8+24x^7y+24xy^7+20x^6y^2+20x^2y^6-24x^5y^3-24x^3y^5-42x^4y^4)=$$ $$=\frac{1}{2}\sum_{cyc}(x^2-y^2)^2(x^4+24x^3y+22x^2y^2+24xy^3+y^4)\geq0$$ and let $x\geq y\geq z$.
Thus, $$\sum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)=$$ $$=\frac{1}{2}\sum_{cyc}(x^5-x^3y^2-x^2y^3+y^5-(x^3z^2-x^2yz^2-y^2xz^2+y^3z^2))=$$ $$=\frac{1}{2}\sum_{cyc}(x-y)^2((x+y)(x^2+xy+y^2)-(x+y)z^2)=$$ $$=\frac{1}{2}\sum_{cyc}(x-y)^2(x+y)(x^2+xy+y^2-z^2)\geq$$ $$\geq\frac{1}{2}\left((x-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)\right)\geq$$ $$\geq\frac{1}{2}\left((y-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)\right)=\frac{1}{2}(y-z)^2(x-y)^2(x+y)\geq0.$$ Done!