Same thing Arturo posted in more detail:
We know that the below two inequalities on the far left are true. So lets use them to prove the one we need to prove$[α(ax_1+(1−a)y_1)+γ(ax_2+(1−a)y_2)≤β]$:
$αx_1+γx_2≤β\;\;|*a\Rightarrow a(\alpha x_1+\gamma x_2)\leq a\beta$
$αy_1+γy_2≤β\;\;|*(1-a)\Rightarrow (1-a)(αy_1+γy_2)\leq (1-a)\beta$
Now add the inequalities on the far right side and we get:
$$a(\alpha x_1+\gamma x_2) + (1-a)(\alpha y_1 + \gamma y_2)\leq a\beta+(1-a)\beta$$
After expanding the expressions in parenthesis on the LHS and rearranging the terms we get:
$$α(ax_1+(1−a)y_1)+γ(ax_2+(1−a)y_2)≤a\beta+(1-a)\beta$$
Which almost looks exactly like the one we need to prove. The RHS after expanding:
$$a\beta+(1-a)\beta=a\beta +\beta -a\beta =\beta \Rightarrow$$
$$\Rightarrow a(\alpha x_1+\gamma x_2) + (1-a)(\alpha y_1 + \gamma y_2)\leq\beta$$
Which is what we needed to show.
2&3 questions:
This set M={$x∈ℝ^2_+∣αx_1+γx_2≤β$} looks like a budget constraint and is bounded by:
if $x_1=0;\;$ $\gamma x_2\leq \beta$;$\;\;x_2\leq \frac{\beta}{\gamma}$
if $x_2=0;\;$ $\alpha x_1\leq \beta$;$\;\;x_1\leq \frac{\beta}{\alpha}$
