Is it true that $$\sup_{x\in\mathbb{R}}\frac{\left\left(1+s\right)+tx\right+\left\left(1+t\right)x+s\right}{1+\leftx\right+\lefts+tx\right}\geq1$$ for all $s,t\in\mathbb{R}$? Is it also true that $$\sup_{x\in\mathbb{R}}\frac{\left\left(1+s\right)x+t\right+\left\left(1+t\right)+sx\right}{1+\leftx\right+\leftsx+t\right}\geq1 \ ?$$ Any answer to either of the two questions would be highly appreciated.

If I denote $$ f(s,t;x)=\frac{\left\left(1+s\right)+tx\right+\left\left(1+t\right)x+s\right}{1+\leftx\right+\lefts+tx\right}, $$ then for $t\ne0$ $$ f\left(s,t;\frac{s}{t}\right)=\frac{1+s/t}{1+s/t}=1, $$ hence the supremum over all $x\in\mathbb R$ is at least 1. If $t=0$, then $$ f(s,0;x)=\frac{1+s+x+s}{1+s+x} $$ and $$ \lim_{x\to\pm\infty}f(s,0;x)=1, $$ so the supremum is (at least) 1 as well. A similar substitution $x=t/s$ works for the second supremum. 

