Show that if the optimal value of the objective function of a linear programming problem is attained at several extreme points, then it is also attained at any convex combination of these extreme points.
Definition. convex combination
A point $x\in\mathbb{R}^n$ is a convex combination of the points $x_1,x_2\dots x_r$ in $\mathbb{R}^n$ if for some real number $c_1,c_2,\dots,c^r$ which satisfy$$\sum_{i=1}^rc_i=1\text{ and }c_i\ge0,\space1\le i\le r$$have$$x=\sum_{i=1}^rc_ix_i$$
Proof.
let $f:\mathbb{R}^m\to\mathbb{R}$ be the objective function, $M$ be the optimal value, $c_1,\dots,c_n\ge0$ and $\sum_{i=1}^nc_i=1$
Assume $x_1,\dots,x_n$ are optimized solutions which are on the extreme points
$$f(x_1),\dots,f(x_n)=M$$
Show any convex combination of $x_1,\dots,x_n,s.t.$
$$x=c_1x_1+\dots+c_nx_n$$ is also a optimized solution$$\text{WTS } f(c_1x_1+\dots+c_nx_n)=M$$
Since $f$ is linear, have the following:
$$f(c_1x_1+\dots+c_nx_n)=c_1f(x_1)+\dots+c_nf(x_n)$$
$$=c_1M+\dots+c_nM=(c_1+\dots+c_n)M=M$$
Therefore convex combination of also also has a optimized value
Now show them are all feasible solutions
Since $x_1,\dots x_n$ are feasible we have
$$Ax_1,\dots,Ax_n\le b\wedge x_1,\dots,x_n\ge0$$
$$\text{WTS }A(c_1x_1+\dots+c_nx_n)\le b\wedge c_1x_1+\dots c_nx_n\ge0$$
Since $Ax_1,\dots,Ax_n\le b$ that
$$c_1Ax_1\le c_1b\wedge\dots\wedge c_nAx_n\le c_nb$$
$$\Rightarrow c_1Ax_1+\dots+c_nAx_n\le b(c_1+\dots+c_n)$$
Since A is a linear transformation, have the following:
$$\boxed{A(c_1x_1+\dots+c_nx_n)\le b}$$
Second part that $$\boxed{c_1x_1+\dots+c_nx_n\ge0}\text{ is trivial}$$
This proves any convex combination of these extreme points are optimized feasible solution.$\tag*{$\square$}$
Is my proof correct, any suggestions$?$
