I found this easier than P1 From basic trigonometry, $\sin \frac{A}{2} = \frac{r - r_1}{r + r_1}$. It suffices to prove that for any $ \triangle ABC$, we have \[ \frac{1 - \sin \frac{A}{2}}{1 + \sin \frac{A}{2}} + \frac{1 - \sin \frac{B}{2}}{1 + \sin \frac{B}{2}} + \frac{1 - \sin \frac{C}{2}}{1 + \sin \frac{C}{2}} \ge 1\]which is equivalent to proving \[ \frac{1}{1 + \sin \frac{A}{2}} + \frac{1}{1 + \sin \frac{B}{2}} + \frac{1}{1 + \sin \frac{C}{2}} \ge 2 \]By CS Engel, it suffices to prove that $\sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2} \le \frac{3}{2}$. But this is trivial since \[ \sum \sin \frac{A}{2} = \sum \sqrt{\frac{1 - \cos A}{2}} = \sum \sqrt{\frac{a^2 - (b - c)^2}{4bc}} = \sum \sqrt{\frac{(a + b - c)(a + c - b)}{4bc}}\le \sum \frac{1}{2} \left( \frac{a + b - c}{2b} + \frac{a + c - b}{2c} \right) = \frac{3}{2} \]

IndoMathXdZ wrote: I found this easier than P1 From basic trigonometry, $\sin \frac{A}{2} = \frac{r - r_1}{r + r_1}$. It suffices to prove that for any $ \triangle ABC$, we have \[ \frac{1 - \sin \frac{A}{2}}{1 + \sin \frac{A}{2}} + \frac{1 - \sin \frac{B}{2}}{1 + \sin \frac{B}{2}} + \frac{1 - \sin \frac{C}{2}}{1 + \sin \frac{C}{2}} \ge 1\]which is equivalent to proving \[ \frac{1}{1 + \sin \frac{A}{2}} + \frac{1}{1 + \sin \frac{B}{2}} + \frac{1}{1 + \sin \frac{C}{2}} \ge 2 \]By CS Engel, it suffices to prove that $\sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2} \le \frac{3}{2}$. But this is trivial since \[ \sum \sin \frac{A}{2} = \sum \sqrt{\frac{1 - \cos A}{2}} = \sum \sqrt{\frac{a^2 - (b - c)^2}{4bc}} = \sum \sqrt{\frac{(a + b - c)(a + c - b)}{4bc}}\le \sum \frac{1}{2} \left( \frac{a + b - c}{2b} + \frac{a + c - b}{2c} \right) = \frac{3}{2} \] Jensen suffices for the last part right?

I dont think there is a need of so much computation/calculation. This is just a direct result by Steiner's Theorem EDIT:Oh ok got it @below

lneis1 wrote: I dont think there is a need of so much computation/calculation. This is just a direct result by Steiner's Theorem There is a difference, the circles are not of the form incircle and excircle as the incircle and the excircle are not tangent to each other but rather of form nine point circle and excircles, I am finding the triangle which satisfies this condition

Ofc trig works, but here is a cuter solution I found in contest. Let $r_a, r_b, r_c, s, \Delta, I$ represent the exradii, the semi-perimeter, the area and the incentre of the triangle ABC. Main Lemma: $\dfrac{r_1}{r} \geq \dfrac{r}{r_a}$ Consider the homothety that sends circle $r_1$ to circle $r$. It sends the first intersection of $AI$ and incircle to the second intersection of $AI$ and the incircle. In particular, this homothety factor is less than the homothety which sends the first intersection of the incircle and AI to the first intersection of $AI$ and the excircle. Since the homotheties are $\dfrac{r}{r_1}$ and $\dfrac{r_a}{r}$, the conclusion follows. $$\sum \dfrac{r_1}{r} \ge \sum \dfrac{r}{r_a} = \sum \dfrac{\dfrac{\Delta}{s}}{\dfrac{\Delta}{s-a}} = \sum \dfrac{s-a}{s} = 1$$

Hello Everyone ))

IndoMathXdZ wrote: I found this easier than P1 Weird, I found it a lot harder than P1. I eventually find a solution though.

This felt harder than P1 for sure. Here is my solution, which seems to be basically the sol in #2, but anyways, here it is. Let the incenter of $ABC$ be $I$, and the incenters of the other three triangles be $I_1,I_2,I_3$. We have that $sin(A/2)=r_1/AI_1$. Thus $r_1=AI_1.sin(A/2)=(AI-r-r_1)sin(A/2)$. Now using $sin(A/2)=r/AI$ two times and bashing gives $r_1=r.(1-sin(A/2)) /(1+sin(A/2))$ and similarly we obtain the other two radii. Let $sin(A/2)=x$ and $y, z$ are defined similarly. Dividing the both sides of the inequality by $r$, we obtain that we have to prove $(1-x)/(1+x)+(1-y)/(1+y)+(1-z)/(1+z)=>1$. Adding $1$ to each fraction and dividing the inequality by two, we see that we have to prove $1/(1+x)+1/(1+y)+1/(1+z)=>2$. Using Titu's Lemma, it's equivalent to $x+y+z<=3/2$. By the formula, $x=\sqrt{(1-cos(A))/2}$ and the others similarly, and combining with $a^2=b^2+c^2-2bc.cos(A)$, we have $x=\sqrt{(2bc-b^2-c^2+a^2)/4bc}=\sqrt{(a-b+c)(a+b-c)/4bc}=1/2.\sqrt{(a-b+c)/c.(a+b-c)/b}$. We obtain similar equations for $y, z$. Now summing all of them and using AM-GM, we obtain what we want.

@below That is exactly my solution

Using trigonometry, it is easy to obtain that $r_1 =\dfrac{1 - \sin \frac{A}{2}}{1 + \sin \frac{A}{2}}r$. So, we can get that the initial inequality is equivalent to proving \[ \dfrac{1 - \sin \frac{A}{2}}{1 + \sin \frac{A}{2}} + \dfrac{1 - \sin \frac{B}{2}}{1 + \sin \frac{B}{2}} + \dfrac{1 - \sin \frac{C}{2}}{1 + \sin \frac{C}{2}} \ge 1\] Now let $f(x)=\dfrac{1-\sin x}{1+\sin x}$. By using simple derivatives, it is easy to get that \[f''(x) = \dfrac{2\sin x (1+\sin x) + 4\cos^2x}{(1+\sin x)^3} \] Which is clearly non-negative when $0 < x < \dfrac{\pi}{2}$. So,the function $f(x)$ is convex over the interval $(0,\dfrac{\pi}{2})$. By using jensen's inequality, \[ \dfrac{1 - \sin \frac{A}{2}}{1 + \sin \frac{A}{2}} + \dfrac{1 - \sin \frac{B}{2}}{1 + \sin \frac{B}{2}} + \dfrac{1 - \sin \frac{C}{2}}{1 + \sin \frac{C}{2}} \ge 3 \left( \dfrac{1 - \sin \frac{A+B+C}{6}}{1 + \sin \frac{A+B+C}{6}} \right ) =1\]Which proves the problem.

I think there is an identity: $\sqrt{r_1\cdot r_2}+\sqrt{r_2\cdot r_3}+\sqrt{r_3\cdot r_1}=r.$ We know that $\frac{r_1}{r}=\cfrac{1 - \sin \cfrac{A}{2}}{1 + \sin \cfrac{A}{2}}=\tan^2\left (\frac{\pi-A}{4}\right)$, and there if we use $\sum\tan\left (\frac{\pi-A}{4}\right)\tan\left (\frac{\pi-B}{4}\right)=1$ as $\sum\frac{\pi-A}{4}=\frac{\pi}{2}$ we will get yhe desired result. Now use AM-GM: $$\sum r_1\ge\sum \sqrt{r_1\cdot r_2}=r.$$

IndoMathXdZ wrote: for any $ \triangle ABC$, we have \[ \frac{1}{1 + \sin \frac{A}{2}} + \frac{1}{1 + \sin \frac{B}{2}} + \frac{1}{1 + \sin \frac{C}{2}} \ge 2 \] In the acute-angled $ \triangle ABC $ the following inequality holds:$$ \frac{1}{1+cosA} + \frac{1}{1+cosB} + \frac{1}{1+cosC} \ge \frac{5R}{s} \ \ ; $$

sqing wrote: IndoMathXdZ wrote: for any $ \triangle ABC$, we have \[ \frac{1}{1 + \sin \frac{A}{2}} + \frac{1}{1 + \sin \frac{B}{2}} + \frac{1}{1 + \sin \frac{C}{2}} \ge 2 \] In the acute-angled $ \triangle ABC $ the following inequality holds:$$ \frac{1}{1+cosA} + \frac{1}{1+cosB} + \frac{1}{1+cosC} \ge \frac{5R}{s} \ \ ; $$ In the any $ \triangle ABC $ the following inequality holds:$$ \frac{1}{1+cosA} + \frac{1}{1+cosB} + \frac{1}{1+cosC} \ge \frac{5R}{s} \ \ ; $$

Exact ratios. And a bit of manipulation: the end result does require some thought. Also, I haven't seen this Solution in the above posts, so here you go: $\color{green} \rule{25cm}{0.5pt}$ $\color{green} \textbf{Ratios ...}$ Let $O_1,O_2,O_3$ be the three centers of the three smaller circles respectively. Then, $AI = AO_1+O_1I$ and its variants give \[ \dfrac{r_1}{\sin{\frac{A}{2}}} + r_1 + r = \dfrac{r}{\sin{\frac{A}{2}}} \]Since it seemed good to work in $\dfrac{r_1}{r}$ instead of just $r_1$, solving for $\dfrac{r_1}{r}$ gives us \[ \dfrac{r_1}{r} = \dfrac{1-\sin{\frac{A}{2}}}{1+\sin{\frac{A}{2}}} \]$\color{blue} \rule{25cm}{1.5pt}$ $\color{blue} \textbf{Expansion.}$ Expanding the cyclic sum \[ \sum_{cyc} \dfrac{1-\sin{\frac{A}{2}}}{1+\sin{\frac{A}{2}}} \geq 1 \]gives us \[ 1-2\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}} \geq \sum_{cyc} \sin{\frac{A}{2}}\sin{\frac{B}{2}} \]after dividing the expression by $2$. As this expression is suspiciously close to this cosine identity, \[ \cos^2{A}+\cos^2{B}+\cos^2{C}+2 \cdot \cos{A}\cos{B}\cos{C} = 1, \: \forall A+B+C = 180^{\circ} \]by replacing $A$ with $ \alpha = 90-A$, we get the identity \[ 1-2\sin{\alpha}\sin{\beta}\sin{\gamma}=\sin^2{\alpha}+\sin^2{\beta}+\sin^2{\gamma}, \: \forall \alpha+\beta+\gamma=90^{\circ} \]To finish, apply the inequality $x^2+y^2+z^2 \geq xy+yz+zx$ to the RHS. Note that $\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90^{\circ}$. We are done. $\blacksquare$ $\blacksquare$ $\blacksquare$

an older source : 1998 Argentina IberoAmerican Training List p39

It's easy to get that $r_1+r_2+r_3\geq r$ $\iff$ $\frac{1}{1 + \sin \frac{A}{2}} + \frac{1}{1 + \sin \frac{B}{2}} + \frac{1}{1 + \sin \frac{C}{2}} \ge 2$. Then use that $\sin \frac{A}{2}=\frac{r}{AI}$ and similarly others. Let $a,b,c$ be length of $BC,AC,AB$,respectivley. And let $a+b-c=x$, $a+c-b=y$ and $b+c-a=z$. So we get that $\sum \frac{1}{1 + \sin \frac{A}{2}}=\sum \frac{\sqrt{(x+y)(x+z)}}{\sqrt{(x+y)(x+z)}+\sqrt{yz}}$. Now we'll prove that $\sum \frac{\sqrt{(x+y)(x+z)}}{\sqrt{(x+y)(x+z)}+\sqrt{yz}}\geq 2$ for any $x,y,z\in\mathbb{R}^+$ $\implies$ $\sum \left(\frac{\sqrt{(x+y)(x+z)}}{\sqrt{(x+y)(x+z)}+\sqrt{yz}}-1\right)\geq -1$ $\implies$ $\sum \frac{\sqrt{yz}}{\sqrt{(x+y)(x+z)}+\sqrt{yz}}\leq 1$ Since $\sqrt{(x+y)(x+z)}\geq \sqrt{xy}+\sqrt{xz}$ we get $\sum \frac{\sqrt{yz}}{\sqrt{(x+y)(x+z)}+\sqrt{yz}}\leq \sum \frac{\sqrt{yz}}{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}=1$ So we are done!

blargh Scale so that $r=1$. If $r_1$ is the circle closest to $A$, etc., then by trig bash we obtain $$r_1=\frac{\cos^2 \frac{A}{2}}{(1+\sin \frac{A}{2})^2}$$and similarly for $r_2$ and $r_3$. Thus we want to prove that $$\sum_{\mathrm{cyc}}\frac{\cos^2 \frac{A}{2}}{(1+\sin \frac{A}{2})^2} \geq 1$$where $A,B,C$ are angles of a triangle, or $$\sum_{\mathrm{cyc}}\frac{\cos^2 A}{(1+\sin A)^2} \geq 1$$given $A+B+C=\pi/2$. The function $f(x)=\frac{\cos^2 x}{(1+\sin x)^2}$ is convex by painfully computing $f''(x)$, so by Jensen's we only have to check the inequality when $A=B=C=\pi/6$, in which case $\frac{\cos^2 A}{(1+\sin A)^2}=\frac{1}{3}$ so we're done. $\blacksquare$

look at this: https://artofproblemsolving.com/community/c6h1252262p6457516

$\textsl{Sin Chasing:}$ The marsh colored segment can be easily expressed with $r_{1}$ and $r$ (green and light blue colored segments) through triangles similarity. So since this segment is $\frac{2r_{1}^2}{r+r_{1}}$, lets check the $sin\, \alpha$ , the $sin\, \alpha$ is $\frac{r-r_{1}}{r+r_{1}}$. $\textsl{Some implementations:}$ The problem asks us to prove that $\frac{r_{1}}{r}+\frac{r_{2}}{r}+\frac{r_{3}}{r}\geq 1$ and it's also easy to notice that: $\frac{r_{1}}{r}$ is $\frac{1-sin\, \alpha }{1+sin\, \alpha }$ so let's replace it in the asked inequality and add $+1$ for every fraction, we get that: $$(!)\sum \frac{1 }{1+sin\, \alpha }\geq 2$$Let's use T2 Lemma, and then prove that $\frac{3}{2}\geq sin\, \alpha +sin\, \beta +sin\, \gamma $ by Jensen's inequality.

The equality $r = \sqrt{r_1r_2} + \sqrt{r_2r_3} + \sqrt{r_3r_1}$ (which trivializes the problem) incidentally is a WOOT homework question, and likely has been for years. A proof would be to simply continue down the path of many of these solutions and take focus off finding a bound.

Let $\omega$ be the incircle of $\triangle ABC$ and $I$ its incenter. $D,E,F$ are the tangency points of $\omega$ with the sides of the triangle. Define $\omega_A $ the circle tangent to $\omega$ and sides $AB$ and $AC$ and similarly define $\omega_b$ and $\omega_c$. First, begin with some new notations: Let $\{ T,R\} =AI\cap\omega $ such that $AT<AR$ and $\{L\}=AI\cap EF$. Note that $\frac{r_1}{r}=\frac{AT}{AR}$(we get this by looking at the homotety that sends $\omega_A$ to $\omega$). Since $(A,L/T,R)=-1$, (because $EF$ is the polar of $A$ with respect to $\omega$) we have that $\frac{AT}{AR}=\frac{LT}{LR}$. All that remains for us to prove is that $\sum \frac{LT}{LR}\ge1$, equivalent to $\sum \frac{2r}{LR}\ge4$. But $LHS=\sum \frac{2r}{\frac{RE^2}{2r}}=\frac{4r^2}{RE^2}\ge\sum\frac{4r^2}{ED^2}$(note that in this sum, $DE, EF, FD$ appear exactly once). So all that remains is to prove that $\sum\frac{1}{ED^2}\ge\frac{1}{r^2}=\frac{16[DEF]^2}{\Pi ED^2}$. So it suffices that $\Pi\space ED^2\cdot EF^2 \ge\sum-ED^4+2\sum\Pi\space ED^2\cdot EF^2$(by Heron's formula), equivalent to $\sum ED^4\ge\Pi\space ED^2\cdot EF^2$, which is obviously true. Equality case for equlateral triangle. We are done.