For positive $a$, $b$, $c$, $d$ with $a+b+c+d=4$, show that $\sum_{cyc} \left(\frac{1}{a^2}-a^2\right)\ge0$

Question

From MOP 2012:

If $a,b,c,d>0$, $a+b+c+d=4$, show $$\sum_{cyc} \left(\frac{1}{a^2}-a^2\right)\ge0$$

My attempt: Using the derivative and tangent line, I obtain $$(x-1)(-x^3+7x^2-x-1)=0$$

Does using $n-1$ EV work?

Answer 1

EV works of course, but there is something nicer: $$\sum_{cyc}\frac{1}{a^2}=\sum_{cyc}\left(\frac{1}{a^2}-\frac{1}{ab}\right)+\sum_{cyc}\frac{1}{ab}=$$ $$=\frac{1}{2}\sum_{cyc}\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\sum_{cyc}\frac{1}{ab}=\frac{1}{2}\sum_{cyc}\left(\frac{1}{a}-\frac{1}{b}\right)^2+\sum_{cyc}\frac{1}{ab}\geq$$ $$\geq\sum_{cyc}\frac{1}{ab}=\left(\frac{1}{ab}+\frac{1}{cd}\right)+\left(\frac{1}{ad}+\frac{1}{bc}\right)$$ and see here: Prove that $\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$