Let $ABC$ be a triangle inscribed in a circle and let \[ l_a = \frac{m_a}{M_a} \ , \ \ l_b = \frac{m_b}{M_b} \ , \ \ l_c = \frac{m_c}{M_c} \ , \] where $m_a$,$m_b$, $m_c$ are the lengths of the angle bisectors (internal to the triangle) and $M_a$, $M_b$, $M_c$ are the lengths of the angle bisectors extended until they meet the circle. Prove that \[ \frac{l_a}{\sin^2 A} + \frac{l_b}{\sin^2 B} + \frac{l_c}{\sin^2 C} \geq 3 \] and that equality holds iff $ABC$ is an equilateral triangle.
Problem
Source: 1997 APMO
Tags: trigonometry, geometry, circumcircle, inequalities, ratio, angle bisector, geometric inequality
darij grinberg
01.01.2005 22:58
One solution can be found on Kalva's site http://www.kalva.demon.co.uk/apmo/asoln/asol973.html , but since you are asking for "another" solution, here is my one: If the angle bisectors of the angles CAB, ABC and BCA meet the sides BC, CA, AB of triangle ABC at the points X, Y, Z and the circumcircle of triangle ABC at the points X', Y', Z' (apart from the points A, B, C, respectively), then you define $l_a=\frac{AX}{AX^{\prime}}$, $l_b=\frac{BY}{BY^{\prime}}$ and $l_c=\frac{CZ}{CZ^{\prime}}$ and ask for the proof of the inequality $\frac{l_a}{\sin^2 A} + \frac{l_b}{\sin^2 B} + \frac{l_c}{\sin^2 C} \geq 3$. In fact, I begin with: Lemma 1. We have $AX\cdot AX^{\prime} = bc$.
Proof of Lemma 1. On the one hand, we have < ACB = < AX'B, since the point X' lies on the circumcircle of triangle ABC. In other words, < ACX = < AX'B. On the other hand, < CAX = < X'AB since the line AX is the angle bisector of the angle CAB. Hence, the triangles ACX and AX'B are similar, and it follows that $\frac{AX}{AC} = \frac{AB}{AX^{\prime}}$, so that $AX\cdot AX^{\prime} = AC\cdot AB$. In other words, $AX\cdot AX^{\prime} = b\cdot c$, and Lemma 1 is proven.
Lemma 2. We have $l_a = 1-\left(\frac{a}{b+c}\right)^2$.
Proof of Lemma 2. Since the angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides, we have $\frac{BX}{CX}=\frac{AB}{AC}$. In other words, $\frac{BX}{CX}=\frac{c}{b}$. This yields
$\frac{BC}{CX} = \frac{BX+CX}{CX} = \frac{BX}{CX}+1 = \frac{c}{b}+1 = \frac{b+c}{b}$,
and thus $\frac{CX}{BC}=\frac{b}{b+c}$, so that
$CX = BC\cdot\frac{b}{b+c}=a\cdot\frac{b}{b+c}=\frac{ab}{b+c}$.
Similarly, $BX = \frac{ac}{b+c}$. Now, after the intersecting chords theorem, $AX\cdot X^{\prime}X = BX\cdot CX$, so that
$AX\cdot X^{\prime}X = \frac{ac}{b+c}\cdot\frac{ab}{b+c} = bc\left(\frac{a}{b+c}\right)^2$,
and thus
$\frac{AX\cdot X^{\prime}X}{bc} = \left(\frac{a}{b+c}\right)^2$.
This shows
$l_a=\frac{AX}{AX^{\prime}}=\frac{AX^{\prime}-X^{\prime}X}{AX^{\prime}}=1-\frac{X^{\prime}X}{AX^{\prime}}=1-\frac{AX\cdot X^{\prime}X}{AX\cdot AX^{\prime}}$.
By Lemma 1, we have $AX\cdot AX^{\prime}=bc$; thus,
$l_a=1-\frac{AX\cdot X^{\prime}X}{bc}=1-\left(\frac{a}{b+c}\right)^2$,
and Lemma 2 is proven.
Lemma 3. For any triangle ABC with sidelengths a, b, c and circumradius R, we have the inequality $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \geq \frac{1}{R^2}$, with equality if and only if the triangle ABC is equilateral.
Proof of Lemma 3. By the extended law of sines, we have a = 2R sin A, b = 2R sin B and c = 2R sin C. Thus, the inequality in question,
$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \geq \frac{1}{R^2}$,
rewrites in the form
$\frac{1}{\left(2R\sin A\right)^2}+\frac{1}{\left(2R\sin B\right)^2}+\frac{1}{\left(2R\sin C\right)^2} \geq \frac{1}{R^2}$.
Multiplying this with $4R^2$, we get
$\frac{1}{\sin^2 A}+\frac{1}{\sin^2 B}+\frac{1}{\sin^2 C} \geq 4$.
But this is true. Indeed, the function $f\left(x\right) = \frac{1}{\sin^2 x}$ is convex on the interval $\left[0;\;\pi\right]$ (since the function $g\left(x\right) = \frac{1}{\sin x}$ is convex on this interval, since the function $h\left(x\right)=x^2$ is convex on whole $\mathbb{R}$, and since $f=g\circ h$). The angles A, B, C of our triangle ABC lie in this interval, and thus we can apply the Jensen inequality and obtain
$\frac{1}{\sin^2 A}+\frac{1}{\sin^2 B}+\frac{1}{\sin^2 C} \geq 3\frac{1}{\sin^2 \frac{A+B+C}{3}} = 3 \frac{1}{\sin^2 \frac{\pi}{3}} = 3 \frac{1}{\left(\frac{\sqrt3}{2}\right)^2} = 4$.
This proves the inequality in question. Because of the strict convexity of our function f(x), equality can hold only for A = B = C, what corresponds to the case when the triangle ABC is equilateral. Lemma 3 is proven.
Lemma 4. For any three positive numbers a, b, c, we have the inequality $\frac{1}{\left(b+c\right)^2}+\frac{1}{\left(c+a\right)^2}+\frac{1}{\left(a+b\right)^2} \leq \frac14 \left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)$.
Proof of Lemma 4. From the QM-HM inequality, applied to the numbers $\frac{1}{b}$ and $\frac{1}{c}$, we have
$\sqrt{\frac12\left(\frac{1}{b^2}+\frac{1}{c^2}\right)} \geq \frac{2}{b+c}$,
with equality only for b = c. Thus,
$\frac12\left(\frac{1}{b^2}+\frac{1}{c^2}\right) \geq \left(\frac{2}{b+c}\right)^2$.
Dividing by 4, we get
$\frac18\left(\frac{1}{b^2}+\frac{1}{c^2}\right) \geq \frac{1}{\left(b+c\right)^2}$.
Remember that equality holds only for b = c here. Similarly, we get
$\frac18\left(\frac{1}{c^2}+\frac{1}{a^2}\right) \geq \frac{1}{\left(c+a\right)^2}$
with equality only for c = a, and
$\frac18\left(\frac{1}{a^2}+\frac{1}{b^2}\right) \geq \frac{1}{\left(a+b\right)^2}$
with equality only for a = b. Adding these three inequalities together, we get
$\frac18\left(\frac{1}{b^2}+\frac{1}{c^2}\right)+\frac18\left(\frac{1}{c^2}+\frac{1}{a^2}\right)+\frac18\left(\frac{1}{a^2}+\frac{1}{b^2}\right)$
$\geq \frac{1}{\left(b+c\right)^2}+\frac{1}{\left(c+a\right)^2}+\frac{1}{\left(a+b\right)^2}$;
in other words,
$\frac14 \left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right) \geq \frac{1}{\left(b+c\right)^2}+\frac{1}{\left(c+a\right)^2}+\frac{1}{\left(a+b\right)^2}$.
Thus,
$\frac{1}{\left(b+c\right)^2}+\frac{1}{\left(c+a\right)^2}+\frac{1}{\left(a+b\right)^2} \leq \frac14 \left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)$.
Equality holds only if all three conditions b = c, c = a and a = b are fulfilled, i. e. if we have a = b = c. This proves Lemma 4.
Lemma 5. For any triangle ABC with sidelengths a, b, c and circumradius R, we have the inequality $\frac{1}{a^2}-\frac{1}{\left(b+c\right)^2}+\frac{1}{b^2}-\frac{1}{\left(c+a\right)^2}+\frac{1}{c^2}-\frac{1}{\left(a+b\right)^2} \geq \frac{3}{4R^2}$, with equality if and only if the triangle ABC is equilateral.
By Lemma 4, we have
$\frac{1}{\left(b+c\right)^2}+\frac{1}{\left(c+a\right)^2}+\frac{1}{\left(a+b\right)^2} \leq \frac14 \left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)$.
Thus,
$\frac{1}{a^2}-\frac{1}{\left(b+c\right)^2}+\frac{1}{b^2}-\frac{1}{\left(c+a\right)^2}+\frac{1}{c^2}-\frac{1}{\left(a+b\right)^2}$
$=\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)-\left(\frac{1}{\left(b+c\right)^2}+\frac{1}{\left(c+a\right)^2}+\frac{1}{\left(a+b\right)^2}\right)$
$\geq \left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)-\frac14\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right) =\frac34\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)$.
On the other hand, from Lemma 3, we have
$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \geq \frac{1}{R^2}$.
Thus,
$\frac{1}{a^2}-\frac{1}{\left(b+c\right)^2}+\frac{1}{b^2}-\frac{1}{\left(c+a\right)^2}+\frac{1}{c^2}-\frac{1}{\left(a+b\right)^2}$
$\geq \frac34\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right) \geq \frac34\cdot \frac{1}{R^2} = \frac{3}{4R^2}$.
Equality holds if and only if equality holds in both Lemma 3 and Lemma 4 - but in Lemma 3, equality holds if and only if the triangle ABC is equilateral, and in Lemma 4, equality holds if and only if a = b = c, what is clearly the same condition. Thus, in our inequality, equality holds if and only if the triangle ABC is equilateral. Thus, Lemma 5 is proven.
Now we can easily solve the problem: After Lemma 5, we have the inequality
$\frac{1}{a^2}-\frac{1}{\left(b+c\right)^2}+\frac{1}{b^2}-\frac{1}{\left(c+a\right)^2}+\frac{1}{c^2}-\frac{1}{\left(a+b\right)^2} \geq \frac{3}{4R^2}$
with equality if and only if the triangle ABC is equilateral. But
$\frac{1}{a^2}-\frac{1}{\left(b+c\right)^2}=\frac{1}{a^2}\left(1-\left(\frac{a}{b+c}\right)^2\right)$.
From Lemma 2, we have $l_a = 1-\left(\frac{a}{b+c}\right)^2$; thus,
$\frac{1}{a^2}-\frac{1}{\left(b+c\right)^2}=\frac{1}{a^2}l_a=\frac{l_a}{a^2}$.
The extended Sine Law yields a = 2R sin A, so that
$\frac{1}{a^2}-\frac{1}{\left(b+c\right)^2}=\frac{l_a}{\left(2R\sin A\right)^2}=\frac{l_a}{4R^2\sin^2 A}$.
Similarly,
$\frac{1}{b^2}-\frac{1}{\left(c+a\right)^2}=\frac{l_b}{4R^2\sin^2 B}$;
$\frac{1}{c^2}-\frac{1}{\left(a+b\right)^2}=\frac{l_c}{4R^2\sin^2 C}$;
thus, our inequality
$\frac{1}{a^2}-\frac{1}{\left(b+c\right)^2}+\frac{1}{b^2}-\frac{1}{\left(c+a\right)^2}+\frac{1}{c^2}-\frac{1}{\left(a+b\right)^2} \geq \frac{3}{4R^2}$
rewrites as
$\frac{l_a}{4R^2\sin^2 A}+\frac{l_b}{4R^2\sin^2 B}+\frac{l_c}{4R^2\sin^2 C} \geq \frac{3}{4R^2}$.
Multiplying this inequality with $4R^2$, we get
$\frac{l_a}{\sin^2 A} + \frac{l_b}{\sin^2 B} + \frac{l_c}{\sin^2 C} \geq 3$.
And, as seen before, equality holds if and only if the triangle ABC is equilateral. This solves the problem.
We can extend our problem: If D, E, F are any three points on the segments BC, CA, AB, and the lines AD, BE and CF intersect the circumcircle of triangle ABC at the points D', E', F', respectively, then, denoting $w_a=\frac{AD}{AD^{\prime}}$, $w_b=\frac{BE}{BE^{\prime}}$ and $w_c=\frac{CF}{CF^{\prime}}$, we have the inequality $\frac{w_a}{\sin^2 A} + \frac{w_b}{\sin^2 B} + \frac{w_c}{\sin^2 C} \geq 3$. Equality holds if and only if the triangle ABC is equilateral and the points D, E, F are the midpoints of its sides BC, CA, AB. This extended version of the problem, of course, yields the original problem in the case when D = X, E = Y and F = Z.
The solution of the extended problem is easy:
Since the point X' lies on the angle bisector of the angle CAB, we have < BAX' = < X'AC. Now, in a circle, equal chordal angles correspond to equal chords; hence, since the chordal angles < BAX' and < X'AC of the chords BX' and CX' are equal, these chords BX' and CX' must be equal, too. Hence, the point X' is the midpoint of the arc BC on the circumcircle of triangle ABC.
Now, among all points on a circular arc, the midpoint of the arc has the greatest distance to the chord bounding the arc. This means that, since the point X' is the midpoint of the arc BC on the circumcircle of triangle ABC, among all points on this arc, the point X' has the greatest distance to the line BC. Hence, since the point D' lies on this arc BC, we have $d\left(X^{\prime};\;BC\right)\geq d\left(D^{\prime};\;BC\right)$, where d(P; g) denotes the distance from a point P to a line g. Equality holds if and only if D' = X', or, equivalently, if and only if D = X.
Now, by Thales, we have $\frac{XX^{\prime}}{AX} = \frac{d\left(X^{\prime};\;BC\right)}{d\left(A;\;BC\right)}$ and $\frac{DD^{\prime}}{AD} = \frac{d\left(D^{\prime};\;BC\right)}{d\left(A;\;BC\right)}$. Since $d\left(X^{\prime};\;BC\right)\geq d\left(D^{\prime};\;BC\right)$, we thus have $\frac{XX^{\prime}}{AX} \geq \frac{DD^{\prime}}{AD}$. Consequently,
$\frac{AX^{\prime}}{AX}=\frac{AX+XX^{\prime}}{AX}=1+\frac{XX^{\prime}}{AX} \geq 1+\frac{DD^{\prime}}{AD} = \frac{AD+DD^{\prime}}{AD}=\frac{AD^{\prime}}{AD}$,
and thus $\frac{AX}{AX^{\prime}} \leq \frac{AD}{AD^{\prime}}$, so $l_a \leq w_a$, and thus $w_a \geq l_a$. Still, as above, equality holds if and only if D = X. Similarly, $w_b \geq l_b$ with equality if and only if E = Y, and $w_c \geq l_c$ with equality if and only if F = Z. Since we also know
$\frac{l_a}{\sin^2 A} + \frac{l_b}{\sin^2 B} + \frac{l_c}{\sin^2 C} \geq 3$
from our problem, and equality here holds if and only if the triangle ABC is equilateral, we thus get
$\frac{w_a}{\sin^2 A} + \frac{w_b}{\sin^2 B} + \frac{w_c}{\sin^2 C} \geq \frac{l_a}{\sin^2 A} + \frac{l_b}{\sin^2 B} + \frac{l_c}{\sin^2 C} \geq 3$.
Equality holds here if and only if it holds in each of the inequalities $w_a \geq l_a$, $w_b \geq l_b$, $w_c \geq l_c$ and $\frac{l_a}{\sin^2 A} + \frac{l_b}{\sin^2 B} + \frac{l_c}{\sin^2 C} \geq 3$. This means that, in the case of equality, we have simultaneously D = X, E = Y, F = Z and the triangle ABC is equilateral. Of course, in an equilateral triangle ABC, the angle bisectors are also the medians, so that the points X, Y, Z are the midpoints of the sides BC, CA, AB, respectively, and the conditions D = X, E = Y, F = Z just mean that the points D, E, F are the midpoints of the sides BC, CA, AB. This completes the solution of the extended version of the problem.
Darij
jgnr
06.07.2008 09:12
Let the bisector of $ BAC$ intersect $ BC$ at $ P$ and the circle at $ Q$. Consider $ \triangle ABP$, by sine rule we have
$ \frac{AP}{AB}=\frac{\sin B}{\sin APB}=\frac{\sin B}{\sin(180^\circ-A/2-B)}=\frac{\sin B}{\sin(A/2+B)}$.
Consider $ \triangle ABQ$. We have $ \angle ABQ=\angle ABC+\angle CBQ=\angle B+\angle CAQ=\angle B+A/2$, $ \angle AQB=\angle ACB=C$. So
$ \frac{AQ}{AB}=\frac{\sin ABQ}{\sin AQB}=\frac{\sin(B+A/2)}{\sin C}$.
So $ l_A=\frac{m_A}{M_A}=\frac{AP}{AQ}=\frac{\sin B\sin C}{\sin^2(A/2+B)}$. Similarly, we have $ l_B=\frac{\sin C\sin A}{\sin^2(B/2+C)}$ and $ l_C=\frac{\sin A\sin B}{\sin^2(C/2+A)}$. Thus,
$ \frac{l_A}{\sin^2A}+\frac{l_B}{\sin^2B}+\frac{l_C}{\sin^2C}\ge\frac{\sin B\sin C}{\sin^2A}+\frac{\sin C\sin A}{\sin^2B}+\frac{\sin A\sin B}{\sin^2C}\ge3$ (by AM-GM).
Equality holds only if $ \sin^2(C/2+A)=1$ and so forth, which implies $ A=B=C=60^\circ$. Hence proved.
panamath
27.02.2013 09:00
This following nice solution is due to a very close friend of mine:
Let $h_i$ be the altitude to side $i$
See that $m_i \ge h_i$ and also $M_i \le 2R$, so;
\[ \frac{l_a}{\sin^2 A}+\frac{l_b}{\sin^2 B}+\frac{l_c}{\sin^2 C}\ge \frac{h_a}{ 2R \sin^2 A}+\frac{h_b}{2R \sin^2 B}+\frac{h_c}{2R \sin^2 C} = \frac{h_a}{ a \sin A}+\frac{h_b}{b \sin B}+\frac{h_c}{c \sin C} = \frac{ah_a}{ a^2 \sin A}+\frac{bh_b}{b^2 \sin B}+\frac{ch_c}{c^2 \sin C} = \frac{bc}{a^2} + \frac{ac}{b^2} + \frac{ab}{c^2} \ge 3 \]
Equality holds when $\frac{bc}{a^2} = \frac{ac}{b^2} = \frac{ab}{c^2}$ $\iff$ $a=b=c$
TheBernuli
27.03.2014 21:30
$\dfrac{l_a}{sin^2A}=\dfrac{4b^2c^2}{(b+c)^2(c+a-b)(a+b-c)}$. By $CS$: $2(\sum ab)^2\geq 3(2\sum a^2b^2+\sum a^2bc-\sum a^4) \Leftrightarrow \sum a^2bc+3\sum a^4\geq 4\sum a^2b^2$. Does anybody know to finish this?
ltf0501
04.03.2016 03:54
TheBernuli wrote: $\dfrac{l_a}{sin^2A}=\dfrac{4b^2c^2}{(b+c)^2(c+a-b)(a+b-c)}$. By $CS$: $2(\sum ab)^2\geq 3(2\sum a^2b^2+\sum a^2bc-\sum a^4) \Leftrightarrow \sum a^2bc+3\sum a^4\geq 4\sum a^2b^2$. Does anybody know to finish this? It can be proved by Schur inequality.
Grandmaster2000
04.03.2016 04:24
How about the law of sines?
BrainiacVR
19.01.2022 11:00
Diagram attached for convenience By some trivial angle chasing, we get $\angle DBC= \angle DAC = \angle BCD= \delta$ Sine rule in $\triangle ABD \Longrightarrow $ $ \frac{M_a}{\sin 2\alpha + \delta}=\frac{c}{\sin2(\alpha + \delta)} \Longleftrightarrow \frac{M_a}{c}=\frac{\sin 2\alpha + \delta}{\sin2(\alpha + \delta)} $ Sine rule in $\triangle ABP \Longrightarrow $ $ \frac{m_a}{\sin 2\alpha }=\frac{c}{\sin2\alpha + \delta} \Longleftrightarrow \frac{m_a}{c}=\frac{\sin 2\alpha }{\sin2\alpha + \delta} $ Equation 2 $\div$ Equation 1 gives $\dashrightarrow$ $ \frac{m_a}{M_a}= \frac{\sin 2\alpha}{\sin2\alpha + \delta} \div \frac{\sin 2\alpha + \delta}{\sin2(\alpha + \delta)}$ $ \Rightarrow l_a=\frac{\sin 2(\alpha + \delta)}{\sin2\alpha + \delta} \times \frac{\sin 2\alpha}{\sin2\alpha + \delta} $ $ \Rightarrow l_a= \frac{\sin B \times \sin C}{\sin^2 \frac{A}{2}+B}$ Similarly, $ l_b= \frac{\sin C \times \sin A}{\sin^2 \frac{B}{2}+C}$ and $ l_c= \frac{\sin A \times \sin B}{\sin^2 \frac{C}{2}+A}$ $ \frac{l_A}{\sin^2A}+\frac{l_B}{\sin^2B}+\frac{l_C}{\sin^2C}=\frac{\sin B\sin C}{\sin^2A \times \sin^2(\frac{A}{2}+B)}+\frac{\sin C\sin A}{\sin^2B \times \sin^2(\frac{B}{2}+C)}+\frac{\sin A\sin B}{\sin^2C \times \sin^2(\frac{C}{2}+A)} $ Now, it's time for some magic We know that $ 0<\sin^2(\frac{C}{2}+A)\leq1$ $ \Rightarrow 0<1 \leq \frac{1}{\sin^2(\frac{C}{2}+A)}$ Applying similar logic for $\sin^2(\frac{B}{2}+C)$ and $\sin^2(\frac{A}{2}+B)$, we can conclude the following $\dashrightarrow$ $ \frac{l_A}{\sin^2A}+\frac{l_B}{\sin^2B}+\frac{l_C}{\sin^2C}\ge\frac{\sin B\sin C}{\sin^2A}+\frac{\sin C\sin A}{\sin^2B}+\frac{\sin A\sin B}{\sin^2C}$ Using AM-GM inequality, we get : $\frac{l_a}{\sin^2A}+\frac{l_b}{\sin^2B}+\frac{l_c}{\sin^2C}$ $\ge 3 \times$$$ \sqrt[3]{ \frac{\sin B\times \sin C\times \sin C\times \sin A\times \sin A\times \sin B}{\sin^2B\times \sin^2C\times \sin^2a}}$$$\Longleftrightarrow \frac{\sin B\sin C}{\sin^2A}+\frac{\sin C\sin A}{\sin^2B}+\frac{\sin A\sin B}{\sin^2C} \ge 3$ Hence Proved For the second part, equality holds iff $ \sin(C/2+A)=$ \sin^2(A/2+B)=$ \sin^2(B/2+C)=1$, which can be easily proved and solved
I literally wasted half an hour on this problem thinking that $m_a=AI$ rather than $AP$
Attachments:
MihaiT
19.01.2022 16:52
BrainiacVR wrote: Diagram attached for convenience By some trivial angle chasing, we get $\angle DBC= \angle DAC = \angle BCD= \delta$ Sine rule in $\triangle ABD \Longrightarrow $ $ \frac{M_a}{\sin 2\alpha + \delta}=\frac{c}{\sin2(\alpha + \delta)} \Longleftrightarrow \frac{M_a}{c}=\frac{\sin 2\alpha + \delta}{\sin2(\alpha + \delta)} $ Sine rule in $\triangle ABP$ $ \frac{m_a}{\sin 2\alpha }=\frac{c}{\sin2\alpha + \delta} \Longleftrightarrow \frac{m_a}{c}=\frac{\sin 2\alpha }{\sin2\alpha + \delta} $ Equation 2 $\div$ Equation 1 gives $\dashrightarrow$ $ \frac{m_a}{M_a}= \frac{\sin 2\alpha}{\sin2\alpha + \delta} \div \frac{\sin 2\alpha + \delta}{\sin2(\alpha + \delta)}$ $ \Rightarrow l_a=\frac{\sin 2(\alpha + \delta)}{\sin2\alpha + \delta} \times \frac{\sin 2\alpha}{\sin2\alpha + \delta} $ $ \Rightarrow l_a=\frac{\sin 2(\alpha + \delta)}{\sin2\alpha + \delta} \times \frac{\sin 2\alpha}{\sin2\alpha + \delta} $ $ \Rightarrow l_a= \frac{\sin B \times \sin C}{\sin^2 \frac{A}{2}+B}$ Similarly, $ l_b= \frac{\sin C \times \sin A}{\sin^2 \frac{B}{2}+C}$ and $ l_c= \frac{\sin A \times \sin B}{\sin^2 \frac{C}{2}+A}$ $ \frac{l_A}{\sin^2A}+\frac{l_B}{\sin^2B}+\frac{l_C}{\sin^2C}=\frac{\sin B\sin C}{\sin^2A \times \sin^2(\frac{A}{2}+B)}+\frac{\sin C\sin A}{\sin^2B \times \sin^2(\frac{B}{2}+C)}+\frac{\sin A\sin B}{\sin^2C \times \sin^2(\frac{C}{2}+A)} $ Now, it's time for some magic We know that $ 0<\sin^2(\frac{C}{2}+A)\leq1$ $ \Rightarrow 0<1 \leq \frac{1}{\sin^2(\frac{C}{2}+A)}$ Applying similar logic for $\sin^2(\frac{B}{2}+C)$ and $\sin^2(\frac{A}{2}+B)$, we can conclude the following $\dashrightarrow$ $ \frac{l_A}{\sin^2A}+\frac{l_B}{\sin^2B}+\frac{l_C}{\sin^2C}\ge\frac{\sin B\sin C}{\sin^2A}+\frac{\sin C\sin A}{\sin^2B}+\frac{\sin A\sin B}{\sin^2C}$ Using AM-GM inequality, we get : $\frac{l_a}{\sin^2A}+\frac{l_b}{\sin^2B}+\frac{l_c}{\sin^2C}$ $\ge 3 \times$$$ \sqrt[3]{ \frac{\sin B\times \sin C\times \sin C\times \sin A\times \sin A\times \sin B}{\sin^2B\times \sin^2C\times \sin^2a}}$$$\Longleftrightarrow \frac{\sin B\sin C}{\sin^2A}+\frac{\sin C\sin A}{\sin^2B}+\frac{\sin A\sin B}{\sin^2C} \ge 3$ Hence Proved For the second part, equality holds iff $ \sin(C/2+A)=$ \sin^2(A/2+B)=$ \sin^2(B/2+C)=1$, which can be easily proved and solved
I literally wasted half an hour on this problem thinking that $m_a=AI$ rather than $AP$
beautiful solution!