The Problem :
If $a, b$ and $c$ are positive constants, and $x$ is a variable, then find the minimum value of the function $$f(x) = \sqrt{a^2+x^2} + \sqrt{(x-b)^2+c^2}$$
I used the following "concept" to solve this problem : Let $P$ be a moving point on a given straight line $l$, and $A$ and $B$ be two fixed points. For the quantity $PA + PB$ to be minimum , the points $A', B$ and $P$ must be collinear, where $A'$ is the image of point $A$ in the line $l$.
Now the function $f(x)$ is also of the form $PA+PB$, where $P \equiv (x,0), A \equiv (0,a)$ and $B \equiv (b,c)$ . The point $P$ moves on the x axis, so the image of $A$ in the x axis is $(0,-a)$. Now using the fact that $A', B$ and $P$ are collinear, I get the desired value of $x$ which yiels the minimum value of the function $f(x)$.
This all is the geometric interpretation of this problem. I was wondering if there is any "algebraic" method to arrive at the answer. Of course, direct use of derivatives for minimization is death here. Maybe some clever use of inequalities may help. I have tried a lot of approaches, like Cauchy-Schwarz Inequality, brute force Calculus, and other inequalities. But nothing seems to help. I am not able to make even the slightest of progress without geometric methods. Any help will be gratefully acknowledged. Thanks in Advance.
You should calculate the derivate of the function you found:
$$f'(x) = {x\over \sqrt{a^2+x^2}} + {x-b\over \sqrt{(x-b)^2+c^2}}$$
and now you find the $x$ by solving $f'(x)=0$:
$$ (b-x)\sqrt{a^2+x^2} = x\sqrt{(b-x)^2+c^2}$$ so $$ (b-x)^2a^2+(b-x)^2x^2 = x^2(b-x)^2 +x^2c^2$$ so $$ (b-x)a = \pm xc\;\;\; \Longrightarrow \;\;\; x = {ab\over a\pm c}$$