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!

If $l+t=a$ then $lt\le\frac{1}{4}a^2$ and we have
$$6lt-(l+t)\le{\textstyle\frac{3}{2}}a^2-a={\textstyle\frac{3}{2}}a(a-{\textstyle\frac{2}{3}})\ .$$
By sketching a graph it is easy to see that for $0\le a\le1$ the right hand side is at most $\frac{1}{2}$.
