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Then write a function exp(x) to calculate (e^x) and use binary search.

Use Maclaurin Series of ln(x).

Count me in too

And,
if ( forum.getInterestedMembers() >= enough && topic.found() )
{
timeToArrange( onlineCamp, me ) = enough; //
}

Here are 50 Functional Equations ( and Inequalities ) from Mathlinks, various contests, own and including basic equations like Cauchy's, Jensen's and D-Alembert's. ( There may be some confusion with the definition of range and co-domain, Wiki says Range $\subseteq$ Co-domain, but somehow not only me...

এই বইয়ে এত বড় ভুল কেমনে করল :O
ওরা লিখসে other cases are similar, কিন্তু $a\le b \le c$ নে, তারপর ওরা যা করসে তা কর, ( মানে ওদের পুরা সমাধানে swap(a,c); ), দেখ সমাধান আসবে না। ওদেরটা ভুল

পোলাপাইন, This is not a symmetric inequality but a cyclic one. So you Can't WLOG assume that $a\ge b\ge c$, instead you can assume WLOG either $a\ge b \ge c$ OR $a\le b\le x$. And, as it is not symmetric, you can't use Muirhead 's inequality.

Denote the statement by $P(x,y)$.
$P(x,0),P(0,x)$ imply that $f$ is a Cauchy equation and $f(x^2)=xf(x)$
Now, $(x+1)(f(x)+f(1))=(x+1)f(x+1)=f((x+1)^2)=f(x^2 + 2x + 1)$
$=xf(x)+2f(x)+f(1)$
So, $f(x)=xf(1)$.

Let $\triangle ABC$ is a triangle with circumradius $R$. Extend $AB$ to $R$, $BC$ to $P$ an $CA$ to $Q$ and let $D,E,F$ be the midpoints of $BC$, $CA$, $AB$ respectively.
Prove that,
\[\frac{PQ}{AB}+\frac{QR}{BC}+\frac{RP}{CA} \ge \frac{PD+QE+RF}{R}\]

What's your new problem? I couldn't get it.