The problem is:
Let $p(x)=x^4-2x^3+ax^2-2x+1$, let a and x be real numbers, find a such that $p(x)\ge0$.
My intent to solve it:
We see that $(x^2-x+1)^2-3x^2+ax^2\ge0$ then $ax^2\ge3x^2-(x^2-x+1)^2$ and $a\ge3-(x-1+{\frac 1x})^2$.
Then the RHS of the inequlity must be as bigger has possible.
It is bigger when $(x-1+{\frac 1x})^2$ is small.
First, it is a square, so it is always positive $x^2\ge0$.
Supose that the minumum of the expression is reached when $x$ is negative, then $(-x-1-1/x)^2=(-1)^2*(x+1+{\frac 1x})^2$ must be smaller than the expression for $x$ positive which is obviously false.
Hence the problem reduces to find the smallest expression of $(x-1+{\frac 1x})^2$.
$x-1+{\frac 1x}\ge1$ beacause of the Arithmetic-Geometric mean inequality.
Therefore the expression inside the square is alwars bigger-equal than $1$ . And $x^2\ge x$ when $x\ge1$.
We conclude that the minimum of $(x-1+{\frac 1x})^2$ is $1$.
Therefore $a\ge3-1=2\ge3-(x-1+{\frac 1x})^2$
$a\ge2$
Is this solution correct?
Since $$p(x)=(x^2-x)^2+(x-1)^2+(a-2)x^2=(x-1)^2(x^2+1)+(a-2)x^2$$ where $(x-1)^2(x^2+1)>0$ for all real number $x$ it follows that $(a-2)x^2$ must be greater or equal to $0$ for all real number $x$, which means $\color{blue}{a\ge 2}$.