Are you smarter than me? 45

$\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b}$

$=\frac{a(a+\frac{1}{a})}{b+c}+\frac{b(b+\frac{1}{b})}{a+c}+\frac{c(c+\frac{1}{c})}{a+b}$

As $a,b,c$ are positive integers, from the Arithmetic Mean & Geometric Mean Relation for a positive real numbers $x$ & $\frac{1}{x}$ :

$x+\frac{1}{x}\geq2$

As it is observed, the minimum value that $x+\frac{1}{x}$ can attain is $2$ , for $x=1$ . This is applicable for $x=a,b,c$ .

Therefore, for the minimum value of the expression, $\boxed{a=b=c=1}$ .

Minimum value of the given expression is

$\Rightarrow \min(\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b})=3*(\frac{1+1}{1+1})=\boxed{3}$