# An Inequality with Angle Trisectors

### Solution 1

Let $\displaystyle m=\tan\alpha.\,$ Then $0\lt m\lt\sqrt{3}.\,$ Choose $A=(0,0,\,$ $B=b,-mb),\,$ $F=(c,mc),\,$ where $b,c\gt 0.\,$ Then $\displaystyle E=\left(\frac{2bc}{b+c},0\right).\,$ We also get that $BC\,$ is described by $mx(b+c)+y(b-c)=2mbc\,$ and $AC\,$ by $2mx=(1-m^2)y.\,$ To find the coordinates of $C,\,$ we solve the two equations:

$\displaystyle x=\frac{2(1-m^2)bc}{(3-m^2)b-(1+m^2)c},\,$ $\displaystyle y=\frac{4mbc}{(3-m^2)b-(1+m^2)c}.\,$

(Note that, since $y\gt 0,\,$ $\displaystyle \frac{b}{c}\gt\frac{1+m^2}{3-m^2}.\,)$ Thus we have,

$\displaystyle C=\left(\frac{2(1-m^2)bc}{(3-m^2)b-(1+m^2)c}, \displaystyle y=\frac{4mbc}{(3-m^2)b-(1+m^2)c}\right).$

Further, $AB=b\sqrt{1+m^2},\,$ $AE=\frac{2bc}{b+c},\,$ $AF=c\sqrt{1+m^2},\,$ $AC=\displaystyle\frac{2(1+m^2)bc}{(3-m^2)b-(1+m^2)c}.\,$ Combining all this, $\displaystyle \frac{AE}{AB}+\frac{AF}{AC}\lt 2\,$ is equivalent to

$\displaystyle \frac{2c}{(b+c)\sqrt{1+m^2}}+\frac{(3-m^2)b-(1+m^2)c}{2b\sqrt{1+m^2}}\lt 2.$

Let $\displaystyle \frac{b}{c}=x,\,$ so that $\displaystyle x\gt\frac{1+m^2}{3-m^2}\gt\frac{1}{3}.\,$ In terms of $x,\,$ the inequality becomes

$\displaystyle \frac{2}{(x+1)\sqrt{1+m^2}}+\frac{(3-m^2)x-(1+m^2)}{2b\sqrt{1+m^2}}\lt 2$

which is equivalent to

$\displaystyle 4x+(x+1)[(3-m^2)x-(1+m^2)]\lt 4x(x+1)\sqrt{1+m^2}.$

But $(3-m^2)x-(1+m^2)\lt 3x-1\,$ and $4x(x+1)\lt 4x(x+1)\sqrt{1+m^2}.\,$ Thus, suffice it to show that

$4x+(x+1)(3x-1)\le 4x(x+1)\,$

which is equivalent to $(x-1)^2\ge 0,\,$ hence, true.

### Solution 2

With the reference to the above diagram,

1. $BF=FG=GC\,$ and $BG=FC.$

2. $DE\parallel FG,\,$ implying $\displaystyle \frac{DE}{FG}=\frac{AD}{AF}\lt 1.\,$ (*)

3. Triangles $ADB\,$ and $CDF\,$ are similar so that $\displaystyle \frac{AD}{AB}=\frac{CD}{CF}.$

Triangles $AEC\,$ and $BEG\,$ are similar so that $\displaystyle \frac{AE}{AC}=\frac{BE}{BG}.$

4. Let $H=C+\overrightarrow{FG},\,$ $FG=CH.$

$\displaystyle\frac{AD}{AB}+\frac{AE}{AC}=\frac{CD}{CF}+\frac{BE}{BG}=\frac{CD+BE}{CF}.$

Thus, suffice it to prove that $CD+BE\lt 2\cdot CF.$

We have $2\cdot CF=CF+GB=HG+GB\ge HB.\,$ And, further,

\displaystyle \begin{align} HB &= HC+CE+BE\\ &=GF+CE+BE\\ &\ge \underbrace{DE+CE}+BE\;\text{(due to (*))}\\ &=CD+BE. \end{align}

It follows that $2\cdot CF\gt CD+BE,\,$ from which $\displaystyle\frac{AD}{AB}+\frac{AE}{AC}\lt 2.$

### Solution 3

Denote $t=(\cos\alpha-\cos 2\alpha)-(1-\cos\alpha)=2\cos\alpha(1-\cos\alpha)\gt 0.$

\displaystyle \begin{align} \frac{AE}{AB}+\frac{AF}{AC}\lt 2\,&\Leftrightarrow\,\frac{\sin B}{\sin(B+\alpha)}+\frac{C}{\sin (C+\alpha)}\lt 2\\ &\Leftrightarrow\,\sin C\sin(B+\alpha)+\sin B\sin (C+\alpha)\\ &\qquad\qquad\qquad\qquad\lt 2\sin (B+\alpha)\sin (C+\alpha)\\ &\Leftrightarrow\,\frac{1}{2}[\cos(B-C+\alpha)+\cos(B-C-\alpha)]+\cos 2\alpha\\ &\qquad\qquad\qquad\qquad\lt \cos(B-C)+\cos\alpha\\ &\Leftrightarrow\,\cos (B-c)\cos\alpha+\cos 2\alpha\lt\cos (B-C)+\cos\alpha\\ &\Leftrightarrow\,t+(1-\cos\alpha)(1+\cos(B-C))\gt 0. \end{align}

### Acknowledgment

I am grateful to Leo Giugiuc for communicating to me the problem originally posted by Saïd Benhamad at the Perles mathematiques facebook page. Solution 1 is by Leo Giugiuc; Solution 2 is by Rachid Moussaoui; Solution 3 is by anonymous author posted by Leo Giugiuc.

Leo Giugiuc has observed that, although the inequality is always strict, $2\,$ is the best possible upper bound for the sum of the two ratios. For example, the sum tends to $2\,$ for an isosceles $\Delta ABC,\,$ $AB=AC,\,$ with $A\,$ moving away from the base $BC.$