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 $H$.

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:
$$\sqrt{x^2 + (1-y)^2} + \sqrt{x^2 + y^2} + \sqrt{2\sqrt{(1-x)^2 + y^2}} \ge \sqrt{5}. \quad (1)$$
Note that this equality is equivalent to
$$\sqrt{(1-y)^2 + x^2} + \sqrt{x^2 + y^2} + \sqrt{(1-x+y)^2 + (1-x-y)^2} \ge \sqrt{5}, \quad (2)$$
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.