Complex Analysis Proofs Let $f$ = $u + iv$ be an entire function satisfying $u(x,y) \geq v(x,y)-1$ for every $(x,y) \in R^2$. Prove that all functions $f, u, v$ are constant. 
Can someone please help me prove this...

Here's an alternative answer. Consider $$(1-i)f = u +iv - iu +v = u + v + i (v-u).$$
Since $v- u \leq 1$ we get that the imaginary part of $f$ lies entirely in the upper half of the complex plane. In particular $(1-i)f$ is an entire function which misses at least two points so by little Picard must be constant. In other words, $f$ is the constant function.
