<s>[INST] Minimum value of $PA+PB$ is If  $P(x,y,z)$ lie on line $\displaystyle \frac{x+2}{2}=\frac{y+7}{2}=\frac{z-2}{1}$ and $A(5,3,4)$ and $B(1,-1,2)$ . Then minimum value of $PA+PB$ is 
what i try
let $\displaystyle \frac{x+2}{2}=\frac{y+7}{2}=\frac{z-2}{1}=\lambda$
Then $P(2\lambda-2,2\lambda-7,\lambda+2)$ 
$PA+PB=\sqrt{(2\lambda-7)^2+(2\lambda-10)^2+(\lambda-2)^2}+\sqrt{(2\lambda-3)^2+(2\lambda-8)^2+\lambda^2)}$
$PA+PB=\sqrt{9\lambda^2-72\lambda+153}+\sqrt{
9\lambda^2-36\lambda+45}$
how do i minimize it b3cause derivative method is very tedious  help me please
 [/INST] Note that
$$
|PA|+|PB|=3\left(\sqrt{(\lambda-4)^2+1}+\sqrt{(\lambda-2)^2+1}\right).
$$ The expression in the parenthesis is equal to the sum of distance between $X=(\lambda,0)$ and $Y=(4,1)$ and distance between $X=(\lambda,0)$ and $Z=(2,-1)$. Then, by the triangle inequality, $|XY|+|XZ|
$ is minimized when $X$ is on the line segment between $Y$ and $Z$. This gives $$|XY|+|XZ|\ge |YZ|=\sqrt{8}=2\sqrt 2$$ hence giving
$$
|PA|+|PB|\ge 6\sqrt{2}.
$$
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