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.
 A: First apply the Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality (from http://tinyurl.com/84o57u4) 
$$\sqrt{\frac{\left(a+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2+\left(b+\frac{1}{d}\right)^2+\left(d+\frac{1}{b}\right)^2}{4}} \geq \frac{\left(a+\frac{1}{c}\right)+\left(c+\frac{1}{a}\right)+\left(b+\frac{1}{d}\right)+\left(d+\frac{1}{b}\right)}{4}$$ 
Square both sides
$$\frac{\left(a+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2+\left(b+\frac{1}{d}\right)^2+\left(d+\frac{1}{b}\right)^2}{4} \geq \frac{\left( (a+b+c+d)+(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}) \right)^2}{16}$$ 
Since we already know the value of $a+b+c+d=16$, apply Arithmetic Mean, Harmonic Mean inequality, i.e.
$$ \frac{a+b+c+d}{4} \geq \frac{4}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}$$
And take it from there, since you said you only need hints.
A: Since $f(x)=x^2+\frac{1}{x^2}$ is a convex function, by Jensen and AM-GM we obtain:
$$\sum_{cyc}\left(a+\frac{1}{c}\right)^2=\sum_{cyc}f(a)+2\sum_{cyc}\frac{a}{c}\geq 4f\left(\frac{a+b+c+d}{4}\right)+8=\frac{289}{4}$$
