If $A,B,C \leq M$, how can we show $\alpha A + \alpha B + (1-2\alpha )C \leq M$? If $A,B,C \leq M$, how can we show 
$$\alpha A + \alpha B + (1-2\alpha )C \leq M,$$
where $0 \leq \alpha \leq 1/2$? What is a counter-example if $\alpha > 1/2$?
My first thought was to try to manipulate it into  
$$\alpha \frac{A-C}{2} + \alpha \frac{B-C}{2} + (1-\alpha)C,$$
but that doesn't seem to be working...

If $\alpha \in [0, \frac{1}{2}]$, we have this 
$$A \le M, ~~B \le M, ~~c \le M \implies \alpha A \le \alpha M, ~~\alpha B \le \alpha M, ~~ (1-2\alpha) C \le (1-2\alpha) M$$
as we are multiplying the inequalities by non-negative numbers.  Summing the three inequalities on the right gives your result.
However $\alpha \le \dfrac{1}{2}$ is not the same as this condition, unless you also specify it is non-negative. 
