I can't think of a Euclidean way of doing it, but I can outline a way of killing it quickly with co-ordinates: Let $Z$ be the origin, let $XY=YZ=XZ=1$, say $BC$ is the $x$-axis, say $\angle XZB = \theta$; then one can work out the equation of $XZ$ in terms of $\theta$. It clearly suffices to prove that the incentre $I$ and $Y$ lie on the same side of the line $XZ$. Then calculate the gradient of $XB$, which gives the gradient of $IB$ (using say, the $\tan 2\theta$ formula). Similarly calculate the gradient of $YC$, which gives the gradient of $IC$. Now we know the equations of $IB$ and $IC$, so solve for the co-ordinates of $I$. Finally, check that $I$ and $Y$ lie on the same side of $XZ$. I can post the calculations if anyone is interested..

Can you post the calculations please? ( In your original post, I think there must be some typo: you say $Z$ the origin and $BC$ x-axis, but $BC$ does not pass through $Z$. )

Due to miquel theorem we have the circumcircles of $BXZ,CXY,AZY$ passes through a common point namely $M$,notice that if the bisector of $B$ meet the circumcircle of $BXZ$ at $F$ then $FZ=FX$,so $F,Y$ lies perpendicular bisector of $ZX$,but $B$ is acute so $\angle ZFX>60$,similar analysis with other two triangles, then we get $I$ must be inside the triangle $XYZ$.

vanu1996 wrote: ... so $\angle ZFX>60$,similar analysis with other two triangles, then we get $I$ must be inside the triangle $XYZ$. Why?

Trigonometry Another not so creative approach but can be done withouth much problems. I will only explain what are the cases to be dealt with and what to use. Denote $d(P,l)$ distance of $P$ from line $l$ Let $XY=YZ=ZX=a$ Clearly if the bisector of $\angle A$ cuts $ZY$ at $D$ we need $d(D,BC)>d(D,AB)$ now we fix $X,Y,Z,A$ and find the maximal $k$ for which $d(D,BC)>k$ and prove that $k\ge d(D,AB)=d(D,AC)$. To do this try to minimize the angle $\angle (DX,BC)$ from both sides. When minimizing it to one side say $\angle BXD$ consider the cases $\angle BZX\ge 90$(this one is easy), $\angle BZX<90$ and $\angle AZD\ge 90$(also not hard) and when both $\angle BZX,\angle AZD$ are acute. For the last case you will need to express $k,d(D,AB)$ in termes of $90-\angle DZA=x$ and $a$ and $r=DZ$ now you get some inequality that you need for $tg x$ but for this you will need that $sin \angle AZD\le sin \angle AZD/sin \angle AYD=r/(a-r)$.

Warning: This is a terrible solution. Abuse of Notation. The term ``segment'' will always denote a line segment (as opposed to a circular segment). Note. Given a line $\ell$, define its argument to be the angle between $\ell$ and the $x$-axis, starting from the $x$-axis and going counterclockwise. Arguments are always in the range $[0, 180^{\circ})$. Let $I$ be the incenter and let $\omega$ be the incircle of triangle $ABC$. We will use the points $X,Y,Z$ and circle $\omega$ to try to locate $A,B,C$. Claim 1. Note that $X,Y,Z$ each lie on the boundary of triangle $ABC$. Since $\omega$ lies within triangle $ABC$, we have that $X,Y,Z$ lie outside or on the boundary of $\omega$. Furthermore, if $\omega$ intersects line $XY$, all of the intersection points must lie within segment $XY$. Otherwise, part of $\omega$ would lie outside triangle $ABC$, contradiction. Claim 2. Let $D,E,F$ be the tangency points of $\omega$ with sides $BC,CA,AB$ respectively. Suppose none of the lines $XY,YZ,ZX$ intersect $\omega$. If $X$ lies on segment $CD$, then it is impossible to find $Z$ on segment $AB$ such that line $XZ$ does not intersect $\omega$. Similarly, if $X$ lies on segment $BD$, then it is impossible to find $Y$ on segment $AC$ such that line $XY$ does not intersect $\omega$. Thus we have a contradiction. We conclude that at least one of the lines $XY, YZ, ZX$ must intersect $\omega$. We return to the main problem. Assume for the sake of contradiction that $I$ lies outside triangle $XYZ$; our goal is now to show that either $ABC$ cannot be acute, $\omega$ cannot be the incircle of triangle $ABC$, or $X,Y,Z$ cannot be contained within segments $BC,CA,AB$ respectively. We can assume without loss of generality that $I, X$ are on opposite sides of line $YZ$. Furthermore, we can assume that line $YZ$ is ``horizontal'' and $X,I$ are ``below'' and ``above'' line $YZ$ respectively. (See Figure 1.) [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.928819800000007, xmax = 39.76706740000004, ymin = -19.4182868, ymax = 7.799332199999999; /* image dimensions */ /* draw figures */ draw(circle((9.92,-1.78), 5.360485052679468)); draw((4.883420800000009,-6.0188436)--(15.061844000000018,-6.244315)); draw((15.061844000000018,-6.244315)--(9.777368439773172,-14.946352361668911)); draw((9.777368439773172,-14.946352361668911)--(4.883420800000009,-6.0188436)); /* dots and labels */ dot((9.92,-1.78),dotstyle); label("$I$", (10.037052800000014,-1.4449952), NE * labelscalefactor); dot((4.883420800000009,-6.0188436),dotstyle); label("$Y$", (5.012261600000009,-5.6967416), NE * labelscalefactor); dot((15.061844000000018,-6.244315),dotstyle); label("$Z$", (15.19068480000002,-5.922213), NE * labelscalefactor); dot((9.777368439773172,-14.946352361668911),linewidth(3.pt) + dotstyle); label("$X$", (9.908212000000013,-14.7478078), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Figure 1. $I,X$ are on opposite sides of line $YZ$. We see that if segment $YZ$ does not intersect $\omega$, we must have line $YZ$ does not intersect $\omega$, so line $YZ$ must be ``below'' $\omega$. This means that segments $ZX,XY$ must not intersect $\omega$, contradicting Claim 2. Thus we must have that segment $YZ$ intersects $\omega$. We see that there are two tangents from $X$ to $\omega$; line $BC$ must be one of these tangents. Let $X_1, X_2$ be the intersections of the two tangents with $\omega$. Case 1. $\omega$ does not intersect lines $XY, XZ$. We have that $\omega$ is contained within the smaller region $\mathcal{R}$ of the plane bounded by rays $XY, XZ$. Thus the rays $XX_1, XX_2$ must also be contained within $\mathcal{R}$; thus line $BC$ intersects $\mathcal{R}$. However, line $BC$ does not intersect triangle $XYZ$, contradiction. Case 2. $\omega$ intersects line $XY$ but does not intersect line $XZ$. (See Figure 2.) [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.928819800000007, xmax = 39.76706740000004, ymin = -19.4182868, ymax = 7.799332199999999; /* image dimensions */ /* draw figures */ draw(circle((9.92,-1.78), 5.360485052679468)); draw((4.207006600000009,-2.1536196)--(16.70456420000002,-2.1214094)); draw((16.70456420000002,-2.1214094)--(10.48368025146099,-12.960716866859293)); draw((10.48368025146099,-12.960716866859293)--(4.207006600000009,-2.1536196)); /* dots and labels */ dot((9.92,-1.78),dotstyle); label("$I$", (10.037052800000014,-1.4449952), NE * labelscalefactor); dot((4.207006600000009,-2.1536196),dotstyle); label("$Y$", (4.3358474000000085,-1.8315176), NE * labelscalefactor); dot((16.70456420000002,-2.1214094),dotstyle); label("$Z$", (16.83340500000002,-1.7993074), NE * labelscalefactor); dot((10.48368025146099,-12.960716866859293),linewidth(3.pt) + dotstyle); label("$X$", (10.616836400000015,-12.7829856), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Figure 2. $\omega$ intersects line $XY$ but does not intersect line $XZ$. Let $P,Q$ be the intersection points of line $XY$ with $\omega$. (We could possibly have $P = Q$.) Let $t_y, t_z$ be the rays from $X$ tangent to $\omega$ such that $t_y$ is outside $\mathcal{R}$ and $t_z$ is inside $\mathcal{R}$. By the above reasoning, we cannot have $t_z$ coincide with line $BC$, so we must have that $t_y$ coincides with line $BC$. Now one of the tangent lines from $Y$ to $\omega$ must be line $AC$; let this tangent line be $u$. Then the intersection of $u$ and $t_y$ must be point $C$, and we can determine $\angle BCA$ by looking at the angle between $u$ and $t_y$. Sub-case 2.1. $u$, $t_y$ are tangent to the same arc $\overarc{PQ}$ of $\omega$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(11cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -0.7191574814425151, xmax = 24.134336132381687, ymin = -12.285790672877546, ymax = 0.6819027906836906; /* image dimensions */ /* draw figures */ draw(circle((9.92,-1.78), 5.360485052679468)); draw((4.434862052742311,-1.9326727164537993)--(16.185425538842992,-1.947699012471831)); draw((16.185425538842992,-1.947699012471831)--(10.297130641716253,-12.11647235220784)); draw((10.297130641716253,-12.11647235220784)--(4.434862052742311,-1.9326727164537993)); draw((xmin, -1.5220791046650244*xmin + 3.5565750355544212)--(xmax, -1.5220791046650244*xmax + 3.5565750355544212)); /* line */ draw((10.297130641716253,-12.11647235220784)--(14.602673637027614,-4.389093254195133)); draw((xmin, -4.030405913580408*xmin + 15.941621526832158)--(xmax, -4.030405913580408*xmax + 15.941621526832158)); /* line */ /* dots and labels */ dot((9.92,-1.78),dotstyle); label("$I$", (9.979565283395956,-1.6321467960931695), NE * labelscalefactor); dot((4.434862052742311,-1.9326727164537993),dotstyle); label("$Y$", (4.494967236814436,-1.782409756273485), NW * labelscalefactor); dot((16.185425538842992,-1.947699012471831),dotstyle); label("$Z$", (16.245530722915117,-1.7974360522915165), NE * labelscalefactor); dot((10.297130641716253,-12.11647235220784),linewidth(3.pt) + dotstyle); label("$X$", (10.355222683846746,-12.030343640571008), NE * labelscalefactor); label("$t_y$", (2.5265224584523023,-0.09946460225395042), NE * labelscalefactor); label("$u$", (4.089257244327585,-0.11449089827198197), NE * labelscalefactor); dot((4.937572906073237,-3.9588015125398126),linewidth(3.pt) + dotstyle); label("$C$", (4.990835005409479,-3.871064902779872), SW * labelscalefactor); dot((6.89408181391885,-6.204773342346854),linewidth(3.pt) + dotstyle); label("$P$", (6.959279783771613,-6.109983009466574), NE * labelscalefactor); dot((4.574048444430357,-2.1744641498317665),linewidth(3.pt) + dotstyle); label("$Q$", (4.630203900976721,-2.0829356766341163), SE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that the horizontal and vertical lines through $C$ are completely contained within angle $BCA$. This is a contradiction. Sub-case 2.2. $u$, $t_y$ are tangent to different arcs $\overarc{PQ}$ of $\omega$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -0.7191574814425151, xmax = 24.134336132381687, ymin = -12.285790672877546, ymax = 0.6819027906836906; /* image dimensions */ /* draw figures */ draw(circle((9.92,-1.78), 5.360485052679468)); draw((4.434862052742311,-1.9326727164537993)--(16.185425538842992,-1.947699012471831)); draw((16.185425538842992,-1.947699012471831)--(10.297130641716253,-12.11647235220784)); draw((10.297130641716253,-12.11647235220784)--(4.434862052742311,-1.9326727164537993)); draw((xmin, -1.5220791046650244*xmin + 3.5565750355544212)--(xmax, -1.5220791046650244*xmax + 3.5565750355544212)); /* line */ draw((10.297130641716253,-12.11647235220784)--(14.602673637027614,-4.389093254195133)); draw((xmin, 5.269265571976335*xmin-25.30113864743315)--(xmax, 5.269265571976335*xmax-25.30113864743315)); /* line */ /* dots and labels */ dot((9.92,-1.78),dotstyle); label("$I$", (9.979565283395956,-1.6321467960931695), NE * labelscalefactor); dot((4.434862052742311,-1.9326727164537993),dotstyle); label("$Y$", (4.494967236814436,-1.782409756273485), NE * labelscalefactor); dot((16.185425538842992,-1.947699012471831),dotstyle); label("$Z$", (16.245530722915117,-1.7974360522915165), NE * labelscalefactor); dot((10.297130641716253,-12.11647235220784),linewidth(3.pt) + dotstyle); label("$X$", (10.355222683846746,-12.030343640571008), NE * labelscalefactor); label("$t_y$", (2.5265224584523023,-0.09946460225395042), NE * labelscalefactor); label("$u$", (4.570098716904595,0.005719469872270494), NE * labelscalefactor); dot((6.89408181391885,-6.204773342346854),linewidth(3.pt) + dotstyle); label("$P$", (6.959279783771613,-6.109983009466574), NE * labelscalefactor); dot((4.574048444430357,-2.1744641498317665),linewidth(3.pt) + dotstyle); label("$Q$", (4.630203900976721,-2.0829356766341163), SE * labelscalefactor); dot((4.2491899700280085,-2.9110282295774117),linewidth(3.pt) + dotstyle); label("$C$", (4.314651684598058,-2.8192241815176624), SW * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We can do a few things to decrease $\angle BCA$. We can dilate triangle $XYZ$ with respect to $X$ until $Y,Z,I$ are collinear; this will decrease $\angle BCA$. Afterwards, we can shrink triangle $XYZ$ with respect to $Y$ until $Z$ lies on $\omega$. Now we wish to show that in the new diagram, $\angle BCA \ge 90^{\circ}$; this will give us the desired contradiction. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -0.7191574814425151, xmax = 24.134336132381687, ymin = -12.285790672877546, ymax = 0.6819027906836906; /* image dimensions */ /* draw figures */ draw(circle((9.92,-1.78), 5.360485052679468)); draw((4.194441316453806,-1.7974360522915165)--(15.283847777761098,-1.782409756273485)); draw((15.283847777761098,-1.782409756273485)--(9.752157701183853,-11.393630612665907)); draw((9.752157701183853,-11.393630612665907)--(4.194441316453806,-1.7974360522915165)); draw((xmin, -1.5467491152852606*xmin + 3.690510683762557)--(xmax, -1.5467491152852606*xmax + 3.690510683762557)); /* line */ draw((xmin, 2.689327019123712*xmin-13.077660414759567)--(xmax, 2.689327019123712*xmax-13.077660414759567)); /* line */ /* dots and labels */ dot((9.92,-1.78),dotstyle); label("$I$", (9.979565283395956,-1.6321467960931695), NE * labelscalefactor); dot((4.194441316453806,-1.7974360522915165),dotstyle); label("$Y$", (4.254546500525932,-1.6471730921112009), NW * labelscalefactor); dot((15.283847777761098,-1.782409756273485),dotstyle); label("$Z$", (15.343952961833224,-1.6321467960931695), NE * labelscalefactor); dot((9.752157701183853,-11.393630612665907),linewidth(3.pt) + dotstyle); label("$X$", (9.814276027197609,-11.309081431705494), NE * labelscalefactor); label("$t_y$", (2.5866276425244283,-0.09946460225395042), NE * labelscalefactor); label("$u$", (4.585125012922626,-0.08443830623591886), NE * labelscalefactor); dot((6.639805421625977,-6.01970795315027),linewidth(3.pt) + dotstyle); label("$P$", (6.703832751465076,-5.929667457250195), NE * labelscalefactor); dot((4.610181019221642,-2.5152702845660646),linewidth(3.pt) + dotstyle); label("$Q$", (4.675282789030816,-2.428540485048842), E * labelscalefactor); dot((3.9584206153229724,-2.4321729009151873),linewidth(3.pt) + dotstyle); label("$C$", (4.014125764237427,-2.3383827089406526), W * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We will use Cartesian coordinates. Assume that the radius of $\omega$ equals 1. Let $I = (0,0), Z = (1,0), Y = (-x, 0)$ for some $x \ge 1$. Since $XY$ needs to intersect $\omega$, we have \[1 \le x \le \frac{2}{\sqrt{3}}.\]We have $X = (\frac{-x+1}{2}, \frac{-x-1}{2}\sqrt{3})$. Let $\theta$ be the argument of $u$ and let $\phi_1$ be the argument of line $XI$. Let $\phi_2$ be the angle between $t_y$ and ray $XI$. In particular, $\phi_1 + \phi_2$ is the argument of $t_y$. It is clear that $\angle BCA = 180^{\circ} - \phi_1 - \phi_2 + \theta$. We wish to show that $\angle BCA \ge 90^{\circ}$; this is equivalent to showing that $\phi_1 + \phi_2 - \theta \le 90^{\circ}$, or \[ 90^{\circ} + \theta \ge \phi_1 + \phi_2. \] Note that $\tan \theta = \frac{1}{\sqrt{x^2 - 1}}$, so \[ \tan (90^{\circ} + \theta) = -\sqrt{x^2 - 1}.\]We have \[ \tan \phi_1 = \frac{(x+1)\sqrt{3}}{x-1}. \]Letting $d = XI$, we see that $\tan \phi_2 = \frac{1}{\sqrt{d^2 - 1}}$, so \[ \tan \phi_2 = \frac{1}{\sqrt{x^2+x}}. \]This yields \[ \tan (\phi_1 + \phi_2) = \frac{(x+1)\sqrt{3}\sqrt{x} + \frac{x-1}{\sqrt{x+1}}}{(x-1)\sqrt{x} - \sqrt{x+1}\sqrt{3}}. \] I claim that the denominator $(x-1)\sqrt{x} - \sqrt{x+1}\sqrt{3}$ is always negative. This is equivalent to showing that $(x-1)^2x < 3(x+1)$ holds for all $x$ in our interval. Expanding, we get \[ f(x) = x^3 - 2x^2 - 2x - 3 < 0. \]To prove this, we note that $f(1) = -6 < 0$. Furthermore, \[f'(x) = 3x^2 - 4x - 2 = 3(x-r_1)(x-r_2) \]where \[r_1 = \frac{2-\sqrt{10}}{3}, r_2 = \frac{2+\sqrt{10}}{3}. \]Note that $r_1 < 0$ and that $r_2 > \frac{5}{3} > \frac{2}{\sqrt{3}}$, so $f'(x) < 0$ for $x$ in our interval. This means that \[f(x) < f(1) < 0 \]for $x$ in our interval. Since $90^{\circ} + \theta, \phi_1 + \phi_2 $ are both in the interval $(90^{\circ}, 180^{\circ}]$, $90^{\circ} + \theta \ge \phi_1 + \phi_2$ is equivalent to \[ \tan (90^{\circ} + \theta) \ge \tan (\phi_1 + \phi_2). \]This in turn is equivalent to \[ -\sqrt{x^2 - 1} \ge \frac{(x+1)\sqrt{3}\sqrt{x} + \frac{x-1}{\sqrt{x+1}}}{(x-1)\sqrt{x} - \sqrt{x+1}\sqrt{3}}. \]This is equivalent to \[ \sqrt{x^2 - 1} \le \frac{(x+1)\sqrt{3}\sqrt{x} + \frac{x-1}{\sqrt{x+1}}}{-(x-1)\sqrt{x} + \sqrt{x+1}\sqrt{3}} \]and after clearing denominators, we get \[ (x+1)\sqrt{3}\sqrt{x-1} - (x-1)\sqrt{x}\sqrt{x^2-1} \le (x+1)\sqrt{3}\sqrt{x} + \frac{x-1}{\sqrt{x+1}}. \]This is true because $(x+1)\sqrt{3}\sqrt{x-1} - (x-1)\sqrt{x}\sqrt{x^2-1} \le (x+1)\sqrt{3}\sqrt{x-1} < (x+1)\sqrt{3}\sqrt{x} \le (x+1)\sqrt{3}\sqrt{x} + \frac{x-1}{\sqrt{x+1}}$. We have our desired contradiction. Case 3. $\omega$ intersects both lines $XY, XZ$. One of the tangents from $X$ to $\omega$ must be line $BC$; proceed as in the previous case. $\blacksquare$

Here's a very clean and mostly synthetic solution By Miquel's Theorem, $(AYZ), (BZX), (CXY)$ meet at some point $P$. Clearly $\angle ABP + \angle ACP = \angle ZXP+\angle YXP = 60^{\circ}$ and similarly for the other vertices. This implies two things: Firstly, $\angle BPC = 180^{\circ} - \angle PBC-\angle PCB = 180^{\circ} - (180^{\circ} - \angle A -60^{\circ}) = 60^{\circ} + \angle A$ and similarly for $\angle CPA, \angle APB$, meaning that $P$ is fixed. Secondly, if $Q$ is the image of $P$ under $\sqrt{bc}$-inversion, then $Q$ lies on the opposite side of $BC$ as $A$ with $\triangle QBC$ equilateral. We'll show $I$ is outside $\triangle AYZ$; similar arguments combined with the fact that $I$ lies inside $\triangle ABC$ will that $I$ is inside $\triangle XYZ$. By performing a $\sqrt{bc}$ inversion, $Y$ and $Z$ map to points $E,F$ on rays $AC,AB$ extended past $C,B$ with $E,Q,F$ collinear; instead of showing $I$ is outside $\triangle AYZ$, we're then left with showing the $A$-excenter $I_A$ is inside $(AEF)$ (because $I,I_A$ are inverses in $\sqrt{bc}$-inversion). Consider any point $J$ on the internal angle bisector of $J$ and any points $U,V$ on $AB,AC$ with $AUJV$ cyclic. For any fixed $J$, I claim $AU+AV$ is always fixed; indeed, let $J_1,J_2$ be the projections of $J$ onto $AB,AC$. It then follows that $\triangle JJ_1U\cong \triangle JJ_2V$, so $J_1U=J_2V\implies AU+AV=AJ_1+AJ_2$. Furthermore, as $J$ moves towards $A$ along the angle bisector, this value of $AJ_1+AJ_2$ obviously decreases. When $J=I_A$, we have $AJ_1+AJ_2 = a+b+c$, so to prove that $I_A$ lies inside $(AEF)$, it's enough to show $AE+AF > a+b+c$ or $CE+BF>a$. Now we're almost done. Let $\angle E=y, \angle F=z, \angle CQE=w, \angle BQF=x$. Since $E,F$ are not within segments $AC,AB$, we must have $0<w,x<120^{\circ}$. Then clearly $w+x=120^{\circ}$. Meanwhile, we have $w+y=\angle ACQ = \angle C+ 60^{\circ}$, so since $ABC$ is acute we have $60^{\circ} < w+y < 150^{\circ}$ and similarly for $x+z$. By the Law of Sines in $\triangle CEQ, \triangle BQF$, we get $CE= \frac{\sin w}{\sin y}CQ,BF=\frac{\sin x}{\sin z}BQ$. Conveniently we have $CQ=BQ=a$, so we just need to show $\frac{\sin w}{\sin y}+\frac{\sin x}{\sin z}>1$. Case 1: $15^{\circ} \le w,x\le 105^{\circ}$. Then $\frac{\sin w}{\sin y}+\frac{\sin x}{\sin z} \ge \sin w + \sin x$. Since $\sin$ is concave on $[0,\pi]$ and $w+x$ is fixed, we have by Karamata that $\sin w +\sin x \ge \sin 15^{\circ} + \sin 105^{\circ} =\frac{\sqrt{6}}{2}>1$, so we're done. Case 2: Suppose case 1 does not hold. WLOG $w<15^{\circ}, x>105^{\circ}$. Then $x+z<150^{\circ}$ yields $z<45^{\circ}$. Since $x<120^{\circ}$, we have $\frac{\sin w}{\sin y}+ \frac{\sin x}{\sin z} > \frac{\sin x}{\sin z}\ge \frac{\sin 120^{\circ}}{\sin 45^{\circ}}>1$, so once again we're done.

Proceed as above until we want to show that $CE+BF > a$. We will prove the more general result that if $BCQ$ is equilateral, and $\ell$ passes through $Q$ but not the interior of the triangle, and $E,F$ lie on $\ell$ so that $E,C$ and $B,F$ are on the same side of $BJ$, then $CE+BF > a$. Suppose $\angle BQC$ is $60$ measured counterclockwise (henceforth CCW). If the angle $\angle(CQ,\ell)$ measured CCW is greater than $90$ then $CE>a$ and we are done. Similarly if $\angle(\ell,QB)$ measured CCW is greater than $90$ then we are done. Now we may assume that $E,F$ are the foot from $C,B$ to $\ell$. Letting $\alpha = \angle(CQ,\ell)$, and $\beta = \angle(\ell,QB)$ we want $\sin\alpha + \sin \beta > 1$ for acute $\alpha, \beta$ summing to $120$. But this is clear from $\sin\alpha+\sin\beta = 2\sin((\alpha+\beta)/2)\cos((\alpha-\beta)/2) = \sqrt 3 \cos((\alpha-\beta)/2) > 1.5 > 1$ since $|\alpha-\beta| < 30$, where all angle measures are in degrees.

Lemma 1. If $0\leq a\leq b\leq 180^{\circ}$, let $I=[a,b]$ then $$\underset{\theta\in I}{\min}\sin\theta=\min\{\sin a,\sin b\}$$Proof. Obvious. $\blacksquare$ Lemma 2. If $\triangle ZYX$ is an equilateral triangle and $A$ is a point lie inside $\angle ZXY$ but outside $\triangle ZYX$, with $\angle AZY>30^{\circ}$. Let $B_1$ be the projection of $X$ on $AZ$. Let $A_1$ be the intersection of the angle bisector of $\angle ZAY$ with $ZY$. Then $$ZB_1A'<\frac{1}{2}\angle ZB_1X=45^{\circ}$$Proof. Let $\omega$ be the circle with diameter $ZX$. Let $M,N$ be the intersection of the perpendicular bisector of $ZX$ and $\omega$, such that $M$ lies between $M$ and $N$.

CASE II: Both $\angle AZY,\angle AYZ$ is at least $30^{\circ}$. Suppose the perpendicular line from $X$ to $AZ$ intersect $AZ,AY$ at $B_1,C_1$, and the perpendicular line from $X$ to $AY$ intersect $AZ,AY$ at $B_2,C_2$. Notice that $B_1,B_2,C_1,C_2$ lies in the extension of either $AZ$ or $AY$. Now, similar to the above case we have $$f(B)\leq \max\{f(B_1),f(B_2)\}$$Moreover, $$f(B_1)<1, f(C_2)<1$$by Lemma 2. Therefore, by Ceva's theorem we have $$f(B_2)=\frac{\sin\angle AB_2A'}{\sin\angle C_2B_2A'}<1$$as well, this implies $f$ is strictly less than 1. $\blacksquare$ This contradiction completes the proof.

Amir Hossein wrote: The vertices $X, Y , Z$ of an equilateral triangle $XYZ$ lie respectively on the sides $BC, CA, AB$ of an acute-angled triangle $ABC.$ Prove that the incenter of triangle $ABC$ lies inside triangle $XYZ.$ Proposed by Nikolay Beluhov, Bulgaria Prove by contradiction: notice incenter is either in triangle AZY or BZX or CYX without using generality of problem we could assume I is in the triangle AZY Now we choose BZX our angel of freedom now define T as intersection of angle bisector of ABC and segment ZY now we could easily calculate ZT/ZX now we define Q as intersection of angle bisector of ACB and segment ZY now we could easily calculate QY/ZX therfore we could calculate QY+ZT/ZX and by bit of calculation we know QY+ZT/ZX>1 therfore I is in the Quadrilateral BZYC with is contradiction