Minimum value of $ \frac{((x-a)^2 + p )^{\frac 12}}{v_1} + \frac{( (x-b)^2 + q )^{\frac12}} {v_2}$

I need to calculate the minimum value of the expression

$ y = \frac{((x-a)^2 + p )^{\frac 12}}{v_1} + \frac{( (x-b)^2 + q )^{\frac12}} {v_2}$

I’m calculating the first derivative , then calculting the corresponding value for $x$ . But the calculation is getting complex.

Here $v_1 , v_2 , p$ and $q$ are positive integers greatest than $1$. I have also tried using $AM >= GM$ . but that too reduces finding maximum of a degree $4$ polynomial

Solutions Collecting From Web of "Minimum value of $ \frac{((x-a)^2 + p )^{\frac 12}}{v_1} + \frac{( (x-b)^2 + q )^{\frac12}} {v_2}$"

It’s derivation of Snell’s Law from Fermat’s Principle

Assume $a<b$,

T(x) &= \frac{\sqrt{(x-a)^2+p}}{v_1}+\frac{\sqrt{(x-b)^2+q}}{v_2} \\
T'(x) &=\frac{x-a}{v_1 \sqrt{(x-a)^2+p}}-
\frac{b-x}{v_2 \sqrt{(x-b)^2+q}} \\
0 &= \frac{\sin \theta_{1}}{v_{1}}-\frac{\sin \theta_{2}}{v_{2}} \\
s &= \sin \theta_{1} \\
\sin \theta_{2} &= \frac{v_{2} s}{v_{1}} \\
b-a &= \sqrt{p} \tan \theta_{1}+\sqrt{q} \tan \theta_{2} \\
b-a &=
\frac{v_{2} s\sqrt{q}}{\sqrt{v_{1}^{2}-v_{2}^2 s^2}}
\quad \cdots \cdots (*) \\
T_{\min} &=
\frac{\sqrt{p}}{v_{1}\cos \theta_{1}}+\frac{\sqrt{q}}{v_{2}\cos \theta_{2}} \\
&= \frac{\sqrt{p}}{v_{1}\sqrt{1-s^{2}}}+
\frac{v_{1} \sqrt{q}}{v_{2}\sqrt{v_{1}^{2}-v_{2}^2 s^2}}

We seldom calculate the exact value of minimal $T$ but condition for minimum instead.

See also the link here.