We proceed with barycentric coordinates where we take $ \triangle{XYZ}$ to be our reference triangle. WLOG let $ X = (1, 0, 0) $ and $ Y = (0, 1, 0) $ and $ Z = (0, 0, 1). $ Let the homothety with ratio $ 4 $ centered at $ G $ take points $ X, Y, Z $ to points $ X', Y', Z'. $ Since $ G = \left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right) $ we find that $ X' = (3, -1, -1) $ and $ Y' = (-1, 3, -1) $ and $ Z' = (-1, -1, 3). $ This lets us easily compute the equations of lines $ Y'Z', Z'X', X'Y' $ to be $ 2x + y + z = 0 $ and $ x + 2y + z = 0 $ and $ x + y + 2z = 0 $ respectively. Assume for the sake of contradiction that points $ A, B, C $ all lie outside $ \triangle{X'Y'Z'}. $ Letting $ A = (r , s, t) $ for some $ r, s, t \in \mathbb{R} $ that satisfy $ r + s + t = 1, $ we must have that $ s, t > 0 $ and $ 2r + s + t < 0. $ Now let $ M = AY \cap X'Y' $ and $ N = AZ \cap X'Z'. $ Since $ C \in AM $ and $ B \in AN $ such that $ M $ is between $ A $ and $ C $ and $ N $ is between $ A $ ad $ B $, we have since $ BC $ passes through $ X $ that line $ MN $ passes through $ \triangle{XYZ}. $ Now it is easy to compute that line $ AY $ has equation $ rz = tx $ and line $ AZ $ has equation $ ry = sx $ so we find that $ M = (r : - 2r - t : t) $ and $ N = (r : s : -2r - s) $ so line $ MN $ has equation $ (3st + 2rs + 2rt + r^2)x + r(r + 2s + t)y + r(r + s + 2t)z = 0. $ Note that $ 3st + 2rs + 2rt + r^2 = 3(st - r^2) + 2r(2r + s + t) < 3(st - r^2) < 3\left(st - \frac{(s + t)^2}{2}\right) < 0 $ and $ r(r + 2s + t) = r(1 + s) < 0 $ and $ r(r + s + 2t) = r(1 + t) < 0 $ which imply that point $ X $ and segment $ YZ $ are on the same side of $ MN. $ This means that $ MN $ does not pass through $ \triangle{XYZ}, $ contradiction!

Dear Wolstenholme this problem is very easy and we can solve it without calculation. My solution: I use the following Key lemma. Lemma: consider $\triangle ABC$ and let $ X, Y, Z$ be arbitrary points on sides $AB, BC, CA $ then among the triangles $ AXZ, BYZ, CXY, XYZ $ area of $\triangle XYZ $ is not minimal.(this is a famous theorem and you can easily prove it by parallel_ inversion) Now Let the hemothety with center $ G $ (centroid of $\triangle XYZ $) with factor $4$ take the points $ X, Y, Z $ to $ X', Y', Z'$ assume for a sake of a contradiction that $\triangle X'Y'Z'$ doesn't cover any of the points $ A, B, C $ then let altitude from $ A$ to $ YZ $ meet $ Y'Z', YZ $ at $ T, S $ from our assumption we have $ AS> TS $ let the feet of $ G, X $ on $ YZ $ $ R, P $ then we have $ TS=3GR=XP $ so $ AS> TS=XP $ so $ S_{AYZ}> S {XYZ} $ and similarly we get $ S_{XYZ}<S_{CYX}, S_{BXZ} $ but it contradicts our lemma.

andria wrote: Dear Wolstenholme this problem is very easy and we can solve it without calculation. My solution: I use the following Key lemma. Lemma: consider $\triangle ABC$ and let $ X, Y, Z$ be arbitrary points on sides $AB, BC, CA $ then among the triangles $ AXZ, BYZ, CXY, XYZ $ area of $\triangle XYZ $ is not minimal.(this is a famous theorem and you can easily prove it by parallel_ inversion) . Can you post any proof or link for the lemma?

Solution for my lemma: Let $\pi$ the plane of the triangle $ABC$ consider a parallel_transformation of $\pi$ to $\pi '$ such that $XYZ$ is equilateral under the inversion.(in this kind of transformations the ratio of areas of two shapes is invariant after inversion if you are not familiar with this kind of inversion you can read yaglom Geometric Transformations III) so we can assume that $XYZ$ is equilateral triangle. Now assume for a sake of a contradiction that the lemma is not true.then let $ S $ the foot of $Y$ on $YZ$ and $ S'$ is reflection of $S$ in the line $ XZ$ now let $YS\cap \odot (\triangle AXZ)=L$ note that $L$ is midpoint of arc $XAZ$ in $\triangle XAZ$ let $L'$ foot of $A$ on $ XZ $ now because $S_{XYZ}<S_{AXZ}$ we get that $YS=SS'<AL'\le LS\longrightarrow \angle XS'Z=60>\angle XLZ=\angle A\longrightarrow 60>\angle A$ similarly we get $60>\angle B,\angle C$ but it is clearly a contradiction. DONE