If  $a+b+c+d=16$, then $(a+\frac{1}{c})^2+(c+\frac{1}{a})^2 + (b+\frac{1}{d})^2 + (d+\frac{1}{b})^2 \geq \frac{289}{4}$ 
If $a,b,c,d$ are positive integers and $a+b+c+d=16$, prove that 
  $$\left(a+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2+\left(b+\frac{1}{d}\right)^2+\left(d+\frac{1}{b}\right)^2 \geq \frac{289}{4}.$$

I know I have to use some inequality, not sure AM GM will work here or Minkowski inequality.  But I only need hints, not a complete solution.  I want to work on it myself.
Please provide only Hints.

Since $f(x)=x^2+\frac{1}{x^2}$ is a convex function, by Jensen and AM-GM we obtain:
$$\sum_{cyc}\left(a+\frac{1}{c}\right)^2=\sum_{cyc}f(a)+2\sum_{cyc}\frac{a}{c}\geq 4f\left(\frac{a+b+c+d}{4}\right)+8=\frac{289}{4}$$
