Let $t$ be an even integer greater than or equal to $6$ and $n$ is an odd natural number. Then is it always true that :
$\lfloor\frac{5-t}{2} \rfloor+1 + \frac {t+n-3}{2} \geq 2$?
I have tried for many values of $t$ and $n=1$ and this is satisfied.
Any argument from anyone is welcome.
You assume that $t$ is an even integer so $t=2k$. Therefore, one has \begin{align*}\lfloor\frac{5-t}{2} \rfloor+1 + \frac {t+n-3}{2}&= \lfloor\frac{5}{2} \rfloor-k+1+ \frac {t+n-3}{2}\\ &=3-k+k+\frac {n-3}{2}\\ &=\frac{n+3}{2}\\ &\geq \frac{1+3}{2}\\ &=2, \end{align*} because since $n$ is a natural odd number, $n\geq 1$. Remark that the inequality is true for all $t\in 2\mathbb{Z}$.