any idea?

Jjesus wrote: any idea? I tried Ptolemy theorem and then Cauchy inequality for left side but couldn't solve the problem. Any other ideas?

RagvaloD wrote: Let $a,b,c,d,x,y$ denote the lengths of the sides $AB, BC,CD,DA$ and the diagonals $AC,BD$ of a cyclic quadrilateral $ABCD$ respectively. Prove that $$(\frac{1}{a}+\frac{1}{c})^2+(\frac{1}{b}+\frac{1}{d})^2 \geq 8 ( \frac{1}{x^2}+\frac{1}{y^2})$$ Invert in the unit circle centered at \(B\). Let \(A', C' , D'\) be the images of \(A, C, D\). Writing Stewart's theorem for \(\Delta A'BC'\) and the cevian \(BD'\) we find \[\frac{y}{x} = \frac{bc+da}{cd+ab}.\]Combining this with Ptolemy's theorem \(xy=ac+bd\) we find \[x^2 = \frac{(ac+bd)(bc+da)}{(ab+cd)}, \qquad y^2 = \frac{(ac+bd)(ab+cd)}{(bc+da)}.\]Thus our inequality becomes \[\left(\frac{a+c}{ac} \right)^2 + \left(\frac{b+d}{bd} \right)^2 \ge \frac{8}{ac+bd}\left(\frac{bc+da}{ab+cd} + \frac{ab+cd}{bc+da}\right)\]which is true by AM-GM: \[LHS \ge \frac{2(a+c)(b+d)}{abcd} \ge \frac{4}{ac+bd}\cdot \frac{ab+cd+bc+da}{\sqrt{abcd}}\ge RHS.\]