10,11 Find the minimum value of the function $f(x)=\sqrt{a^{2}+x^{2}}+\sqrt{(b-x)^{2}+c^{2}}(a>0, b>0, c>0)$.
Consider the segment $D E$ of length $b$. Construct perpendiculars to this segment at its ends: $D A=a, E B=c$ (points $A$ and $B$ lie in different half-planes relative to $D E$, see the figure). Let point $C$ lie on segment $D E$ and $D C=x$, then $E C=b-x$. From the right triangles $A C D$ and $B C E$ we get:

$A C=\sqrt{a^{2}+x^{2}}, \quad B C=\sqrt{(b-x)^{2}+c^{2}}$.

Since $A C+B C \geq A B$, the minimum value of this function is $A B=\sqrt{(a+c)^{2}+b^{2}}$ (it is achieved when point $C$ lies on segment $A B$.

## Answer

$\sqrt{(a+c)^{2}+b^{2}}$