Selection and Training Session

First solution. Consider the rotation around point $C$ by $90^{\circ }$ such that the images of point $A$ and point $B$ lie in the same half-plane with respect to the line $CB$ (see the Fig.). Denote the images of points $A$ and $X$ by $D$ and $Y$ respectively. By construction, the triangles $ACD$ and $CXY$ are isosceles right triangles with right angles at vertex $C$, hence $AD=\sqrt{2}AC=2AB$ and $XY=\sqrt{2}CX$. From the right triangle $BAD$ one can easily find $BD=\sqrt{5}AB$. Finally, $DY=AX$, since $DY$ is the image of $AX$ under rotation.



The required inequality is equivalent to $AX+XY+YD\ge BD$ which is true, as the length of the polyline $BXYD$ is greater or equal to the length of the segment $BD$. The equality holds if and only if points $X$ and $Y$ lie on the segment $AD$. It is easy to see that there is only one such point. Indeed, let $H$ be the foot of the perpendicular from point $C$ to the line $BD$. Point $H$ is the midpoint of the hypotenuse of an isosceles right triangle $CXY$, therefore the triangle $CHX$ is an isosceles right triangle with a right angle at vertex $H$. Thus, there is exactly one such point $X$, namely, the image of point $C$ (lying inside the triangle $ABC$) under the rotation by $90^{\circ }$ with respect to the point $H$.



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Alternative solution.

Consider points $A(0,1)$, $B(0,0)$, $C(1,0)$ and $X(x,y)$ on the Cartesian plane. Then the required inequality is equivalent to: $$\sqrt{x^2+(1-y{)}^2}+\sqrt{x^2+y^2}+\sqrt{2\sqrt{(1-x{)}^2+y^2}}\ge \sqrt{5}.(1)$$ Note that this equality is equivalent to $$\sqrt{(1-y{)}^2+x^2}+\sqrt{x^2+y^2}+\sqrt{(1-x+y{)}^2+(1-x-y{)}^2}\ge \sqrt{5},(2)$$ since $2(1-x{)}^2+2y^2=(1-x+y{)}^2+(1-x-y{)}^2$. Now the equality (2) follows from $|\vec{a}|+|\vec{b}|+|\vec{c}|\ge |\vec{a}+\vec{b}+\vec{c}|$ for vectors $\vec{a}(1-y,x)$, $\vec{b}(x,y)$, $\vec{c}(1-x+y,1-x-y)$, and $\vec{a}+\vec{b}+\vec{c}=(2,1)$. For any point $X$ the equality in (2) means that vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ are collinear. Hence $(1-y)y=x^2$ and $(1-x-y)x=(1-x+y)y$, which leads to $y-y^2=x^2$ and $y+y^2=x-x^2$. Adding this equations we get $2y=x$ and after the substitution find $x=2/5$, $y=1/5$. It is easy to check that point $X(2/5,1/5)$ satisfy the conditions of the problem.