# 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.

• I'm assuming that $BM_2$ and $CM_3$ are also angle bisectors so that their (and $AM_1$) intersection creates the incenter of the triangle. – Jared May 8 '14 at 7:13
• @Jared Yes, of course. That's what I meant by and so on. – Sawarnik May 8 '14 at 7:25

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.

• I have an alternative, infact stronger and easier, proof of Lemma 1: $$1+\frac{b}{a}+\frac{c}{a}+1+\frac{a}{b}+\frac{c}{b}+1+\frac{a}{c}+\frac{b}{c} \ge 9$$ $$\frac1{a}+\frac1{b}+\frac1{c}\ge \frac9{a+b+c}$$ $$\frac{2A}{a}+\frac{2A}{b}+\frac{2A}{c} \ge \frac{18rs}{a+b+c}$$ $$9r \le h_a+h_b+h_c \le \sum_{cyc} AL_A$$ – Sawarnik May 18 '14 at 15:22
• (+1) Purely Awesome !! :D .. $\sum_{cyc}AL_A \geq 9r$ I started that way too and was stuck trying to show $\ge 2R - r$ .. the Japanese Carnot Theorem .. mind blowing !! :) – r9m May 18 '14 at 17:31

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.