 Solution

The feasible triangle has verticesThese come from the three pairs of active boundary lines and satisfy the remaining inequalities. Its denominator $3x_1+x_2+2$ is positive at each vertex, with minimum $12/5$, so is positive throughout the triangle because it is an affine function.

For a direct linear programming optimality certificate, add twice the second inequality to the third to get $-x_1-3x_2\le 2$. Thus $x_1+3x_2+2\ge 0$, equivalently $2(x_1-x_2)\le 3x_1+x_2+2$. Division by the positive denominator gives an objective at most $1/2$. Both inequalities used in the bound are equalities at $v_3$, where the first inequality is also satisfied. ThereforeEquality requires the two bounding inequalities to be tight, so this optimizer is unique.