Show that $\frac{1}{MA^2}+\frac{1}{MB^2}+\frac{1}{MC^2}+\frac{1}{MD^2}\geq2$ for $M$ inside square $ABCD$ with side length $2$ 
Show that $$\frac{1}{MA^2}+\frac{1}{MB^2}+\frac{1}{MC^2}+\frac{1}{MD^2}\geq2$$ for any point $M$ inside of the square $ABCD$ whose side length is $2$.

I could manage to prove this using analytic geometry. Looking for a geometrical proof and possibly generalization of that.
For an analytic geometry proof, see @MichaelRosenberg's answer to the question Prove:$\frac{1}{(x-1)^2+(y-1)^2}+\frac{1}{(x+1)^2+(y-1)^2}+\frac{1}{(x-1)^2+(y+1)^2}+\frac{1}{(x+1)^2+(y+1)^2}\geq2 $ ,if $-1<x,y<1$ .


COMMENT:
An experimental approach;
In a Pythagorean triple (a, b,c) it can be seen that  $c^2<2.5 ab$.Now considering the sketch we have:
$MA^2<2.5(S_{AHME} =AE \times AH)$
$MB^2<2.5(S_{HBGM} =AE \times BH)$
$MC^2<2.5(S_{EMFC} =AH \times EC)$
$MD^2<2.5 (S_{MGDF}=BH \times EC)$
$\frac{1}{MA^2}+\frac{1}{MB^2}+\frac{1}{MC^2}+\frac{1}{MD^2}>\frac{EC\times BH+AH\times AH\times EC+AE\times BH+AE\times AH}{2.5(AE\times AH\times EC\times BH)}=\frac{AF(AH+BH)+EC(AH+BH)}{2.5\times S_{AHME}\times S_{MGDF}}=\frac{4}{2.5 S_{AHME}\times S_{MGDF}}=\frac{1.6}{S_{AHME}\times S_{MGDF}}$
The denominator $S_{AHME}\times S_{MGDF}$ is fractional, that is  it's value is less than unity; therefore the sum of fractions can be greater than 2. If M is at center of square, then all fractions are equal to $1$ and inequality changes to equality.
