just an idea! have you thought of a way to make a connection between the following inequality that is always true for a,b,c >0 a/(b+c)+b/(c+a)+c/(a+b)>=3/2 perhaps this is a good idea perhaps this ain't such a good one! perhaps it can never be applied to this! it is you who diciede! tschuess!

galois wrote: well this is an old problem lying since quite sum time but i think now i have a solution using lagrangia's idea.i will post it later well i think i met this problem be4 too and i used ptolemy then it simplifies to lagrangia's inequality..

Yeah... that inequality works. I'll just post the quick solution since no one else has bothered. In quadrilateral ABCE we have by Ptolemy's inequality $CE \cdot AB + AE \cdot BC \ge AC \cdot BE$ $\implies BC \cdot (CE + AE) \ge AC \cdot BE$ $\implies \frac{BC}{BE} \ge \frac{AC}{CE+AE}$ Adding the analogous inequalities, $\frac{BC}{BE} + \frac{DE}{DA} + \frac{FA}{FC} \ge \frac{AC}{CE+AE} + \frac{CE}{EA+CA} + \frac{EA}{AC+EC} \ge \frac{3}{2}$

Applying Ptolemy's inequality on $ACDE$ we have $(DE)(AC)+(DC)(AE)\ge (DA)(CE)$. Since $DE=DC$, this becomes $DE(AC+AE)\ge (DA)(CE)$ or $\frac{DE}{DA}\ge\frac{CE}{AC+AE}$. Similarly we have \[\begin{cases}\frac{DE}{DA}\ge\frac{CE}{AC+AE} \\ \frac{FA}{FC}\ge \frac{EA}{CE+CA}\\\frac{BC}{BE}\ge\frac{EA}{EA+EC}\end{cases}\] After adding the three inequalities, the result immediately follows from Nesbitt.

This is a nice application of Ptolemy's inequality.Try another which is from the same. Prove that in an acute-angled triangle ABC with orthocentre H, we have: $AH+BH+CH\le2h_{max}$ where $h_{max}$ denotes the largest altitude of triangle ABC .

What is Ptolemy's inequality?

WJ.JamshiD wrote: What is Ptolemy's inequality? The Ptolemy's Inequality states that given a quadrilateral $ ABCD $ the following is true :- $ AB. CD + AD. BC \ge AC.BD $, with equality iff $ ABCD $ is cyclic.

biomathematics wrote: WJ.JamshiD wrote: What is Ptolemy's inequality? The Ptolemy's Inequality states that given a quadrilateral $ ABCD $ the following is true :- $ AB. CD + AD. BC \ge AC.BD $, with equality iff $ ABCD $ is cyclic. thanks you

Relatively straightforward, only needed Ptolemy's, and considering that I just made like 10 posts on inequalities Nesbitt's was a very easy application to apply and see.

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