<s>[INST] Find the minimum value of $y = \sqrt {{x^2} + 4x + 13} + \sqrt {{x^2} - 8x + 41} $ Find the minimum value of $y = \sqrt {{x^2} + 4x + 13}  + \sqrt {{x^2} - 8x + 41} $ .
My approach is as follow
$y = \sqrt {{x^2} + 4x + 4 + 9}  + \sqrt {{x^2} - 8x + 16 + 25}  \Rightarrow y = \sqrt {{{\left( {x + 2} \right)}^2} + 9}  + \sqrt {{{\left( {x - 4} \right)}^2} + 25} $
Within the square roots the minimum values are 3 and 5 hence the minimum value should be greeter than 8 but not able to find the actual minimum value.
 [/INST] I'll approach as geometry.
Let three point as $A(-2,-3), P(x,0), B(4,5)$. Then $y=\overline{AP}+\overline{PB}$.
By triangle inequality, $y\ge\overline{AB}=10$.
So the minimum value is $10$.
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