How to prove this geometry inequality (1) with $2(DF+EF)\ge BC$ There is a picture which hope to illustrate the configuration:
$\triangle ABC$ is such that $\angle A=\dfrac{2\pi}{3}$, and $F$ is the midpoint of $BC$, and $D,E$ lie on $AB,AC$ respectively, such that $DE ||BC$.
Show that:
$$2(DF+EF)\ge BC$$
maybe use cosine  theorem,let $\dfrac{BD}{AB}=k,AB=c,BC=a,AC=b$, and in $\Delta ABC$ we use cosine theorem we have
$$a^2=b^2+c^2+bc\tag{1}$$
and $$DF^2=BD^2+BF^2-2BD\cdot BF\cos{\angle B}
=(kc)^2+\dfrac{a^2}{4}-2kc\cdot\dfrac{a}{2}\dfrac{a^2+c^2-b^2}{2ac}=k^2c^2+\dfrac{a^2}{4}-\dfrac{k(a^2+c^2-b^2)}{2}$$
the same as
$$EF^2=\dfrac{a^2}{4}+(kb)^2-\dfrac{k(a^2+b^2-c^2)}{2}$$
it is equivalent
$$\sqrt{4k^2c^2+a^2-2k(a^2+c^2-b^2)}+\sqrt{4k^2b^2+a^2-2k(a^2+b^2-c^2)}\ge a,0\le k\le 1\tag{2}$$
I have tried some inequality (AM-GM,Cauchy-Schwarz,$\cdots$) to prove $(2)$, but didn't see the solution.

This is not a complete solution, just some simplification.
Square both sides of (2) and using (1) to get rid of $a^2$, we obtain:
$$2\sqrt{c^2+b^2(1-2k)^2+bc(1-2k)}\sqrt{b^2+c^2(1-2k)^2+bc(1-2k)}+((b^2+c^2)(1-2k)^2+2bc(1-2k))\ge bc\tag{3}$$
Let $\kappa=1-2k$, thus $-1\le \kappa \le 1$. Then (3) becomes
$$2\sqrt{c^2+b^2\kappa^2+bc\kappa}\sqrt{b^2+c^2\kappa^2+bc\kappa}+((b^2+c^2)\kappa^2+2bc\kappa)\ge bc\tag{4}$$
