Let $a$ and $b$ be positive real numbers. Find the minimum possible value of $\sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2} .$

Enter your answer to 2 decimal places.

The answer is 2.82842712474619.

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Let ABCD be a square with $A = (0, 0), B = (1, 0), C = (1, 1), D = (0, 1)$ , and P be a point in the same plane as ABCD. Then the desired expression is equivalent to $AP + BP + CP + DP$ .

By the triangle inequality, $AP + CP \ge AC$ and $BP + DP ≥ BD$ , adding the two inequalities we get the minimum possible value of the expression as $AC + BD = 2\sqrt{2} \approx 2.82$ .

Equality holds if and only if $P=(\frac{1}{2},\frac{1}{2}$ ) or $a=b=\frac{1}{2}$ .