Here's a procedure that is guaranteed to find such proof.
Consider a polynomial function
$$
f(x):=\sum_{k=0}^{n}{a_kx^k}
$$
such that $f(x)\geqslant0$ holds for all $x\in\mathbb{R}$ . It's apparent that when $x$ approaches infinity (no matter $+$ or $-$) , the limit of any polynomial function is either $+\infty$ or $-\infty$ . Here $f(x)\geqslant0$ , thus
$$
\lim_{x\to+\infty}f(x)=\lim_{x\to-\infty}f(x)=+\infty
$$
and by the Rolle's Theorem, there exists at least one stationary point in $\mathbb{R}$ . Since $f$ is of degree $n$ , by the Fundamental Theorem of Algebra, the number of stationary points are finitely many (at most $n-1$).
Consider a finite set $X$ such that $X=\{f(\eta)~|~f^\prime(\eta)=0\}$ , since this set is finite, there exists the smallest elements $\min X=\{f(\xi_1),f(\xi_2),\ldots,f(\xi_p)\}$ . It follows that $f(\xi_1)=f(\xi_2)=\cdots=f(\xi_p)\leqslant f(x)$ for all $x\in\mathbb{R}$ and the equality occurs if and only if $x=\xi_1\vee x=\xi_2\vee\cdots\vee x=\xi_p$ .
Let $g(x)=f(x)-f(\xi_p)$ , note that $g(\xi_1)=g(\xi_2)=\cdots=g(\xi_p)=0$ , thus again by the Fundamental Theorem of Algebra, we can write $g(x)$ in such manner that
$$
g(x)=\left[\prod_{t=1}^{p}{(x-\xi_t)^{m_t}}\right]\left(\sum_{k=0}^{n-m}{b_kx^k}\right)
$$
where $m=\sum_{t=1}^{p}{m_t}$ , and $m_t$ is the largest possible exponent of $(x-\xi_t)$ . It's clear that $m_t$ is always even, else either $f(\xi_t+\varepsilon)<0$ or $f(\xi_t-\varepsilon)<0$ , where $\varepsilon>0$ .
Now we can replicate this procedure for
$$
h(x):=\sum_{k=0}^{n-m}{b_kx^k}
$$
It is obvious that the degree of the polynomial function is strictly decreasing, thus such procedure will end after finitely many iterations.
Because we can write
$$
f(x)=\left[\prod_{t=1}^{p}{(x-\xi_t)^{m_t}}\right]h(x)+\xi_p
$$
The proof of your polynomial $f(x)\geqslant0$ is constructed.
For example, take $f(x)=x^4+x^3+x+2$ , the procedure outputs
$$
f(x)=(x+1)^2\left[(x-\frac12)^2+\frac34\right]+1\geqslant0
$$
