How do we prove $x^6+3x^3+2x^2+x+1 \geq 0$ Question
How do we prove
$$x^6 + 3x^3+2x^2+x+1\geq0$$
My progress
$$x^6+3x^3+2x^2+x+1=(x+1)^2(x^4-2x^3+3x^2-x+1)$$
I appreciate your interest

$$x^6 + 3x^3+2x^2+x+1= (x + 1)^2 \left( {\left( {x - \tfrac{1}{2}} \right)^4  + \tfrac{1}{2}\left( {x - \tfrac{1}{2}} \right)^2  + x^2  + \tfrac{{13}}{{16}}} \right) \ge 0$$
