Sorry for the unrelated comment, but could you please post the rest of the problems from the first two days of Iranian TSTs (unless they are from ISL, of course)?

a_507_bc wrote: Sorry for the unrelated comment, but could you please post the rest of the problems from the first two days of Iranian TSTs (unless they are from ISL, of course)? Yes of course. I will do it soon.

Seemed to easy so probably a fake solve: By taking p to be infinitely small, we get that there exist x, such that both x and $f(x)$ are infinitely small, as well as z such that both z and $f(f(z))$ are infinitely small. By taking such x and z and putting it into the triangle condition we get:\newline 1. $x+f(y) \leq f(f(y)) + f(x)$\newline 2. $f(f(y)) + z \leq f(y) + f(f(z))$\newline by taking x and z such that $x, f(x), z, f(f(z))$ approach zero, we get $f(y) \leq f(f(y))$ from 1. and $f(f(y)) \leq f(y)$ from 2.,which is enough to imply:\newline 3.$f(y) = f(f(y))$\newline With this, the 1. becomes $x \leq f(x)$. Also, the triangle condition becomes $x + z + 2f(y) \geq f(z) + f(x)$ so we only need to take $f(y)$ to get $x + z \geq f(x) + f(z)$ which together with $f(x) \geq x$ implies $f(x) = x $ for all x.

AHZOLFAGHARI wrote: Find all function f:R+ -> R+ such that for every three real positive number $x,y,z$ : $$ x+f(y) , f(f(y)) + z , f(f(z))+f(x) $$ are length of three sides of a triangle and for every postive number $p$ , there is a triangle with these sides and perimeter $p$. Actually , the correct version of this problem has that condition on the area of a triangle , not its perimeter.

Let $p_n$ be any sequence of positive numbers such that $p_n \to 0^+$ as $n \to \infty$. Let $x_n, y_n, z_n$ be such that $$ x_n+f(y_n) + f(f(y_n)) + z_n + f(f(z_n))+f(x_n) = p_n$$Since $f$ gets only positive values, we see that $x_n$, $f(x_n)$, $f(y_n)$, $f(f(y_n))$, $z_n$ and $ f(f(z_n))$ all approach $0$ as $n\to \infty$. Using triangle inequality we have $$ x+f(y) + f(f(y)) + z > f(f(z))+f(x) $$$$f(f(y)) + z + f(f(z))+f(x)> x+f(y)$$Plugging $(y, z) = (y_n, z_n)$ into these inequalities and letting $n \to \infty$ we get $x \geq f(x)$ from the first one and $f(x) \geq x$ from the second one. So $f(x) = x$ for all $x > 0$.

any idea for the correct version?