How can I simplify $2^{x+y-1} < 2^x + 2^y - 1$? I am trying to understand for what values of $x$ and $y$ the following inequality is true:
$2^{x+y-1} < 2^x + 2^y - 1$
where $x$ and $y$ are both non-negative integers.
The expression looks simple, but I'm not sure how to proceed. I checked numerically and it seems like it is true if and only if $x \leq 1$ or $y \leq 1$. But I would like to know if there is a way to get this result without resorting to brute force.
I tried rearranging it like this, so that each side of the inequality can be viewed as a binary number with either 1 or 2 nonzero bits:
$2^{x+y} + 2 < 2^{x+1} + 2^{y+1}$
This made it a little easier to check some special cases, but didn't really get me any closer to an answer.
I would be very interested to know how I could approach this problem and other similar problems. Many thanks :)
 A: Rewrite it like this $$2^x(2^{y-1}-1)< 2^y-1$$
If $y=0$ or $y=1$ then all $x$ are good and if $y\geq 2$ then $$2^x <{2^y-1\over 2^{y-1}-1}\le3 \implies x\in \{0,1\}$$
A: $2^{x+y-1} < 2^x + 2^y - 1$
$2^{x+y} < 2(2^x + 2^y) - 2$
$2^x2^y < 2(2^x + 2^y) - 2$
Let $u = 2^x, v = 2^y$
$uv < 2u + 2v - 2$
$uv - 2u - 2v + 4< 2$
$(u-2)(v-2) < 2$
$(2^x - 2)(2^y-2)< 2$
It would be easy enough to solve this over the reals.
Nonetheless, over the non-negative integers:
If $x = 0$ or $x = 1$ the inequality it true for all y.
And due to the symmetry of the equation, we can swap the x's and y's in the previous conclusion.
A: Let $u=2^x$ and $v=2^y$ then
$$\begin{align}
2^{x+y-1} < 2^x + 2^y - 1 \\
&\iff \frac{uv}2<u+v-1 \\
&\iff 2u-uv+2v-2>0 \\
&\iff u(2-v)>2-2v
\end{align}$$
that is

*

*for $v>2 \iff y>1$
$$u<\frac{2(v-1)}{v-2} \iff x<\log_2 \left(\frac{2(2^y-1)}{2^y-2}\right)$$

*

*for $1\le v<2 \iff 0 \le y<1$
$$u>0\ge \frac{2(1-v)}{2-v}\iff \forall x\in \mathbb R$$

*

*for $0<v<1 \iff y<0 $
$$u>\frac{2(1-v)}{2-v} \iff x>\log_2 \left(\frac{2(1-2^y)}{1-2^y}\right)$$
