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...
 A: 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. 
A: Here is a partial solution that relies on $0\leq \alpha\leq 1/2$:
First, $A\leq M$ hence we have $A-C\leq M-C$. Similarly $B-C\leq M-C$. This gives $ A+B-2C\leq 2M-2C$ multiplying by $\alpha$ and adding $C$ gives
$\alpha A + \alpha B + (1-2\alpha) C \leq 2 \alpha M + (1-2\alpha) C$
So it suffices to show that $2 \alpha M + (1-2\alpha) C\leq M$ but this is immediate because $\alpha\leq 1/2$ so $2\alpha -1\leq 0$ then $C\leq M$ gives
$(2\alpha-1)C\geq (2\alpha-1)M$ rearranging gives $2\alpha M + (1-2\alpha)C\leq M$
A: $$
\color{#ff0000}{\large\alpha A + \alpha B + \left(1 - 2\alpha\right)C}
=
\alpha\ \overbrace{\left(A - M\right)}^{\leq\ 0}
+
\alpha\ \overbrace{\left(B - M\right)}^{\leq 0}
+
\left(1 - 2\alpha\right)\ \overbrace{\left(C - M\right)}^{\leq\ 0}\ +\ M
\color{#ff0000}{\large\color{#0000ff}{\ \leq\ } M}
$$
which is true when $\alpha \geq 0$ and $\left(1 - 2\alpha\right) \geq 0$ which is equivalent to $0 \leq \alpha \leq 1/2$. Otherwise, see the above counterexample ( comment ) by @peterwhy . However, when $A = B = C$, the problem statement is true for any value of $\alpha \in {\mathbb R}$.  
