Simple inequalities Suppose $l,t\in[0,1]$ and $l+t\leq1$ I want to prove $1+l+t>6lt$. When $t=0$ or $l=0$, it is trivial, so I started with $l,t\neq0$ but I couldn't reach anywhere. I don't have time to write in detail what I have already tried, but I tried to manipulate $(l-t)^2$ mostly. Anyway, if anyone help me with the proof that would be great. Many thanks!

I think your problem is possible to state it as a optimization problem:
$$\min 1+l+t-6lt$$
$$s.t.$$
$$ l+t\leq 1$$
$$0\leq l\leq1$$
$$0\leq t\leq1$$
Which have a global minimum at $(l,t)=\left(\frac{1}{2},\frac{1}{2}\right)$ then $$1+l+t-6lt\geq1+\frac{1}{2}+\frac{1}{2}-2\frac{1}{2}\frac{1}{2}=\frac{1}{2}>0$$
Hope it helps
