A geometric inequality, proving $8r+2R\le AM_1+BM_2+CM_3\le 6R$ 
Here, $AM_1$ is the angle bisector of $\angle A$ extended to the circumcircle and so on. $R$ is the circumradius and $r$ is the inradius, respectively. I have to prove that:
$$8r+2R\le AM_1+BM_2+CM_3\le 6R$$
The second part is easy, since each of $AM_1$ is a chord, $AM_1\le 2R$, so $\sum AM_1 \le 6R$. But the first part is giving me nightmares. Applying Euler's inequality gives $8r+2R\le 6R$, which is not much helpful and I'm out of ideas.
Please help. Besides, playing GeoGebra tells that its true.
 A: The quadrilaretal $ABM_1C$ is cyclic. Thus the Ptolemy Theorem says:
$AB.CM_1+AC.BM_1=BC.AM_1$, that is $AM_1=\dfrac{b+c}{a}.BM_1$ 
(Since, angle subtended by equal chords at the circumcircle are equal and vice-versa, $BM_1=CM_1$, as $\angle BAM_1=\angle CAM_1=\dfrac{A}{2}$).
Also, $BM_1=2R\sin\frac{A}{2}$.
Similarly, getting expressions for $BM_2$ and $CM_2$, we have:
$\displaystyle AM_1+BM_2+CM_3=\sum\limits_{cyc} \frac{b+c}{a}.\left(2R\sin\frac{A}{2}\right)=\sum\limits_{cyc} \frac{\sin B+\sin C}{\sin A}.\left(2R\sin\frac{A}{2}\right)$
$\displaystyle =2R\sum\limits_{cyc} \frac{\sin B+\sin C}{\sin A}.\sin\frac{A}{2}=2R\sum\limits_{cyc} \frac{2\sin \frac{B+C}{2}.\cos\frac{B-C}{2}}{2\sin \frac{A}{2}.\cos \frac{A}{2}}.\sin \frac{A}{2}$
$\displaystyle = 2R\sum\limits_{cyc}\cos\frac{B-C}{2}$.
We also have, $r=4R\prod\limits_{cyc}\sin\dfrac{A}{2}=R(-1+\sum\limits_{cyc}\cos A)$, 
Therefore, $8r+2R = 2R(-3+4\sum\limits_{cyc}\cos A)$
Thus, the inequality in the question is, $\displaystyle \sum\limits_{cyc}\cos\frac{B-C}{2} \ge -3+4\sum\limits_{cyc}\cos A = 1 + \frac{4r}{R}$
Using the substitution, $a=x+y$, $b=y+z$ and $c=x+z:$
$\displaystyle \sum\limits_{cyc}\cos\frac{B-C}{2} \ge -3+4\sum\limits_{cyc}\cos A$
$\displaystyle \Leftrightarrow \sum\limits_{cyc} \frac{\sin A + \sin B}{2\cos \frac{C}{2}} \ge \sum\limits_{cyc} \frac{2(b^2+c^2-a^2)}{bc} - 3$
$\displaystyle \Leftrightarrow \sum\limits_{cyc} \frac{\frac{2\Delta}{bc} + \frac{2\Delta}{ac}}{2\sqrt{\frac{1+\frac{a^2+b^2-c^2}{2ab}}{2}}} \ge \sum\limits_{cyc} \frac{2ab^2+2ac^2-2a^3-abc}{abc}$
$\displaystyle \Leftrightarrow \sum\limits_{cyc} \frac{2\Delta(a+b)}{2\sqrt{\frac{(a+b+c)(a+b-c)}{4ab}}} \ge \sum\limits_{cyc} 2ab^2+2ac^2-2a^3-abc$
$\displaystyle \Leftrightarrow \sum\limits_{cyc} \frac{4\Delta(a+b)\sqrt{ab}}{\sqrt{(a+b+c)(a+b-c)}} \ge 2\sum\limits_{cyc} 2ab^2+2ac^2-2a^3-abc$
$\displaystyle \Leftrightarrow \sum\limits_{cyc} \frac{(a+b)\sqrt{ab}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{\sqrt{(a+b+c)(a+b-c)}} \ge 2\sum\limits_{cyc} (2ab^2+2ac^2-2a^3-abc)$
$\displaystyle \Leftrightarrow \sum\limits_{cyc} (a+b)\sqrt{ab(a+c-b)(b+c-a)} \ge 2\sum\limits_{cyc} 2ab^2+2ac^2-2a^3-abc$
$\displaystyle \Leftrightarrow 2\sum\limits_{cyc} (x+z+2y)\sqrt{xz(x+y)(z+y)} \ge 2\sum\limits_{cyc} (xz(x+z) + 6xyz)$
$$\displaystyle \Leftrightarrow \sum\limits_{cyc} (x+z+2y)\sqrt{xz(x+y)(z+y)} \ge \sum\limits_{cyc} (x^2z + xz^2 + 6xyz)$$
On the other hand we have the inequality $\sqrt{(x+y)(z+y)} \ge (y+\sqrt{xz})$ (By Squaring and applying AM-GM)
Thus it suffices to show $\displaystyle \sum\limits_{cyc} (x+z+2y)(y\sqrt{xz}+xz) \ge \sum\limits_{cyc} (x^2z + xz^2 + 6xyz)$
$\displaystyle \Leftrightarrow (\sum\limits_{cyc} xz(x+z)) + 6xyz + (\sum\limits_{cyc} xy\sqrt{xz}+yz\sqrt{xz}+2y^2\sqrt{xz})\ge \sum\limits_{cyc} (x^2z + xz^2) + 18xyz$
$\displaystyle \Leftrightarrow \sum\limits_{cyc} xy\sqrt{xz}+yz\sqrt{xz}+2y^2\sqrt{xz}\ge 12xyz$
Which is AM-GM Inequality with $12$-terms.
A: I have a (way) shorter proof, that relies on the following lemma:
Lemma 1: If $AL_A,BL_B,CL_C$ are the angle bisectors of $ABC$, then:
$$\sum_{cyc}AL_A \geq 9r.$$
Due to the Van Obel's theorem and the angle bisector theorem we know that if $I$ is the incenter of $ABC$, then:
$$\frac{AI}{IL_A}=\frac{b+c}{a}.$$
Since $IL_A\geq r$, we have:
$$\sum_{cyc}AL_A \geq r\cdot\sum_{cyc}\frac{a+b+c}{a}=r\cdot\left(\sum_{cyc}a\right)\cdot\left(\sum_{cyc}\frac{1}{a}\right)$$
hence the claim follows from the Cauchy-Schwarz inequality.

Let $A'B'C'$ be the triangle for which $A'B'$ is the perpendicular to $OC$ through $C$ and so on. We have that the circumcircle of $ABC$ is the incircle of $A'B'C'$; moreover, $M_A$ is the incenter of $A'BC$ and so on. Let $r_A$ be the inradius of $A'BC$. Due to the Japanese-Carnot theorem we know that:
$$r_A+r_B+r_C = 2R-r.$$
Now we have:
$$\sum_{cyc}AM_A = \sum_{cyc}AL_A + \sum_{cyc}L_A M_A \geq \sum_{cyc}AL_A + \sum_{cyc} r_A \geq (9r) + (2R-r) = 8r+2R,$$
QED.
