The least value of $f(x)$=$\sqrt{x^2-2x+2}+\sqrt{x^2-4x+29}$ If $m$ is the least value of $f(x)$=$\sqrt{x^2-2x+2}+\sqrt{x^2-4x+29}$, occur at $x = α$ , then $[m]+[α]$ is equal to (where [.] denotes greatest integer function)
(A)6
(B)7
(C)5
(D)4
My approach is as follow
$f(x)=\sqrt{(x-1)^2+1}+\sqrt{(x-2)^2+5^2}$
$\alpha$ lies between 1&2 and $m$ lies between 6 and 7 hence the answer is 7. I checked it and ur us correct. But solution is required

By using the Cauchy Schwarz inequality, you can easily obtain:
$$\sqrt {a^2+b^2}+\sqrt{c^2+d^2}\ge \sqrt {(a+c)^2+(b+d)^2};$$
and equality holds if $\frac{a}{b}=\frac {c}{d}$. Now, take $a=x-1$, $c=2-x$, $b=1$, $d=5$. We get:
$$\sqrt {(x-1)^2+1^2}+\sqrt{(2-x)^2+5^2}\ge \sqrt {1^2+6^2}=\sqrt {37}\implies m=\sqrt {37}.$$
And,
$$\frac {x-1}{1}=\frac {2-x}{5}\implies a=\frac {7}{6}.$$
