<s>[INST] Mistake in solving this simple inequality? Suppose have the the inequality, where $x,y$ are variables,
$$
\frac{x+y}{2}\leq \frac{\alpha}{\beta+\gamma}(x+y)
$$
Clearly, this is true if $\frac12 <\frac{\alpha}{\beta+\gamma}$, but if I try to solve it by adding/subtracting I get 
$$
x(1-\frac{2\alpha}{\beta+\gamma})\leq y(\frac{2\alpha}{\beta+\gamma} -1) = -y(1-\frac{2\alpha}{\beta+\gamma})
$$
which gives, (assuming I am not dividing by a negative number, if are then flip the inequality sign )
$$
x\leq -y
$$
which is very different from $\frac12 <\frac{\alpha}{\beta+\gamma}$?
Why do these two methods give very different results (or are they same and I am overlooking something?)
 [/INST] Method 2:
$x(1-\frac{2\alpha}{\beta+\gamma})\leq y(\frac{2\alpha}{\beta+\gamma} -1) = -y(1-\frac{2\alpha}{\beta+\gamma})$
Case 1:  $1 = \frac {2\alpha}{\beta + \gamma}$
Then $0 = 0$ and nothing can be determined.
Case 2: $1 > \frac {2\alpha}{\beta + \gamma}$
Then $x\le -y$
Case 3: $1 < \frac {2\alpha}{\beta + \gamma}$
Then $x \ge -y$.
Method 1:
$\frac{x+y}{2}\leq \frac{\alpha}{\beta+\gamma}(x+y)$
$(x+y) = \leq \frac{2\alpha}{\beta+\gamma}(x+y)$
Case 1:  $1 = \frac {2\alpha}{\beta + \gamma}$
The $x + y = x+y$ and nothing can be determined.
Case 2: $1 > \frac {2\alpha}{\beta + \gamma}$
The if $x+y > 0$ we would have $x+1 >  \frac {2\alpha}{\beta + \gamma}(x+y)$ but that is a contradiction.
So $x + y \le 0$ and $x \le -y$.
Case 3: $1 < \frac {2\alpha}{\beta + \gamma}$
If $x + y < 0$ we would have $x+1 >  \frac {2\alpha}{\beta + \gamma}(x+y)$ but that is a contradiction.
So we have $x + y \ge 0$ and $x \ge -y$.
...
so both methods come to same conclusion.
</s>