All TST problems are from ISL2019 , so don't post them

Kamran011 wrote: All TST problems are from ISL2019 , so don't post them This problem is not in ISL2019. That is why I posted it

Let $a_1 \leq a_2 \leq... \leq a_n$ are weights. From condition follows $a_1+...+a_{n-2}>a_n$ Let $s_i=a_1+...+a_i$ Let there are not good triples. Then $a_i>a_{i-1}+a_{i-2}, 3 \leq i \leq n$ $s_{n-2}>a_n>a_{n-1}+a_{n-2} \to s_{n-3}>a_{n-1}>a_{n-2}+a_{n-3} \to s_{n-4}>a_{n-2} \to ... \to s_2>a_4 \to a_1+a_2>a_3 \to $ contradiction

Similar solution. Let $a_1 \leq a_2 \leq ... \leq a_n$ and $S=a_1+a_2+...+a_n$ $a_1,...,a_{n-2},a_n$ is good so we have $a_n<a_1+a_2+...+a_{n-2}=S-a_n-a_{n-1}$ $\implies \boxed{2a_n+a_{n-1}<S}$ Assume that there is no good triplet. $x\leq y\leq z \implies a_x<a_y+a_z$ and $a_y<a_x+a_z$ so $a_z \geq a_x+a_y.$ We have $a_n \geq a_{n-1}+a_{n-2}$ $a_{n-1} \geq a_{n-2}+a_{n-3}$ $...$ $a_3 \geq a_{2}+a_{1}$ By adding these inequalities we get $a_n \geq (a_{n-2}+a_{n-3}+...+a_1)+a_2=S-a_n-a_{n-1}+a_2$ $\implies 2a_n+a_{n-1} \geq S+a_2$ $\implies S>2a_n+a_{n-1} \geq S+a_2$ Contradiction