Doubt: $a$,$b$,$c$ $\in$ $\mathbb{N}$ such that sum of any two of them is greater than the third.
$Prove$ $that$:
$[1-(b-c)\frac 1a ]^a$$[1-(c-a)\frac 1b ]^b$$[1-(a-b)\frac 1c ]^c$$\leq 1$
Thank you.
So it looked like triangular inequality:
What i tried:
$WLG$ $a<b<c$ and then i tried performing triangular inequality or the property that:
$Difference$ $of$ $two$ $sides$ $of$ $a$ $triangle$ $is$ $less$ $than$ $third$ $side$
Issue is: 2 of three terms is less than 1 but one is less than $2^a$. So i was stuck.
Another thing i tried was to take log but i got nowhere too.
Thank you