If $x+y>\theta$, then there are $\alpha,\beta$ with $\alpha+\beta=\theta$, $x>\alpha$ and $y>\beta$

Question

Let $x,y\in[0,\infty]$ and $\theta\in\mathbb R$ with $x+y>\theta$.

How can we show that there are $\alpha,\beta\in\mathbb R$ with $\alpha+\beta=\theta$, $x>\alpha$ and $y>\beta$?

While the problem seems to be trivial, I don't see how we can prove it. Clearly, if $x=y$, then we can choose $\alpha=\beta=\theta/2$.

Answer 1

$x+y > \theta$,

Hence $\exists \epsilon >0$, $x+y - \epsilon = \theta$.

$(x-0.5\epsilon) + (y-0.5\epsilon) = \theta$.

Pick $\alpha = x-0.5\epsilon$ and $\beta = y-0.5\epsilon$.