It's exceptional .What is the source ?

leonardg wrote: It's exceptional .What is the source ? http://treinamentoconesul.blogspot.com.br Lista-I-2010

Thank you . In my solution,I use as lemma a problem of mine accepted for publicaton in a math journal . All I can say for now is equality holds if and only if AD and BC are parallel .

Let $a=\overline{AB}, b=\overline{BC}, c=\overline{CD}, d=\overline{DA}$. Using \begin{align*} \overline{AC} &= \sqrt{\frac{(ac+bd)(ad+bc)}{ab+cd}} \\ \overline{BD} &= \sqrt{\frac{(ac+bd)(ab+cd)}{ad+bc}} \end{align*}we need to prove that $$\sqrt{ac+bd}\left|\sqrt{\frac{ad+bc}{ab+cd}}-\sqrt{\frac{ab+cd}{ad+bc}}\right| \leqslant |a-c|.$$After squaring we get \begin{align*} (ac+bd)\left(\frac{ad+bc}{ab+cd}+\frac{ab+cd}{ad+bc}-2\right) &\leqslant (a-c)^2 \\ \Leftrightarrow \frac{ac+bd}{(ab+cd)(ad+bc)}(a-c)^2(b-d)^2 &\leqslant (a-c)^2. \end{align*}So for $a \neq c$ it suffices to prove that \begin{align*} \frac{ac+bd}{(ab+cd)(ad+bc)}(b-d)^2 &\leqslant 1 \\ \Leftrightarrow 0 &\leqslant bd(a-b+c+d)(a+b+c-d), \end{align*}which is obvious.

pablock wrote: Let $ABCD$ be a cyclic quadrilateral. Show that $\vert \overline{AC} - \overline{BD} \vert \le \vert \overline{AB}-\overline{CD} \vert$ and determine when does equality hold. Obvious or not .I don't know how it feels .

Attachments:

I used the word obvious because the latter inequality is folklore . I first met it in '97,at a Romanian exam for teachers . Mister Chikaya informed me the following: 1998 Romanian Mathematical Olympiad ^^ If ABCD is a cyclic quadrilateral, prove that | AC - BD | ≤ | AB - CD|. When does equality hold ? Proposed by D. Mihet ^^

DerJan wrote: Let $a=\overline{AB}, b=\overline{BC}, c=\overline{CD}, d=\overline{DA}$. Using \begin{align*} \overline{AC} &= \sqrt{\frac{(ac+bd)(ad+bc)}{ab+cd}} \\ \overline{BD} &= \sqrt{\frac{(ac+bd)(ab+cd)}{ad+bc}} \end{align*}we need to prove that $$\sqrt{ac+bd}\left|\sqrt{\frac{ad+bc}{ab+cd}}-\sqrt{\frac{ab+cd}{ad+bc}}\right| \leqslant |a-c|.$$After squaring we get \begin{align*} (ac+bd)\left(\frac{ad+bc}{ab+cd}+\frac{ab+cd}{ad+bc}-2\right) &\leqslant (a-c)^2 \\ \Leftrightarrow \frac{ac+bd}{(ab+cd)(ad+bc)}(a-c)^2(b-d)^2 &\leqslant (a-c)^2. \end{align*}So for $a \neq c$ it suffices to prove that \begin{align*} \frac{ac+bd}{(ab+cd)(ad+bc)}(b-d)^2 &\leqslant 1 \\ \Leftrightarrow 0 &\leqslant bd(a-b+c+d)(a+b+c-d), \end{align*}which is obvious. Why does \begin{align*} \overline{AC} &= \sqrt{\frac{(ac+bd)(ad+bc)}{ab+cd}} \\ \overline{BD} &= \sqrt{\frac{(ac+bd)(ab+cd)}{ad+bc}} \end{align*}?

by the general ptomeys theorem.