<s>[INST] Proof: if $x^5-4x^4+3x^3-x^2+3x-4≥0$ then $x≥0$ I have proved this question by finding the solution of the polynomial using 17 steps Newthon Rhapson and found out that x ≥ 3,0735. I am wondering is there any more simple way to solve this?
 [/INST] You have that
$$
\begin{gathered}
  p(x) = x^5  - 4x^4  + 3x^3  - x^2  + 3x - 4 =  \hfill \\
   \hfill \\
   = x^3 \left( {x^2  - 4x + 3} \right) - \left( {x^2  - 3x + 4} \right) \hfill \\ 
\end{gathered} 
$$
Now:


*

*$$-\left( {x^2  - 3x + 4} \right)<0 \;\;\forall x\;\; \in \mathbb R$$

*$$
{x^2  - 4x + 3}>0\;\; \forall x\;\;<0$$

*$$x^3<0 \forall x\;\;<0$$
Therefore, for every x<0 you have that p(x)<0. Hence, if p(x)=0 it must be x>0


In order to prove that if p(x)=0 then x>3 we have that 


*

*If $$ 0< x <1 $$ then $$ f(x) > 0 $$

*If $$ 1< x <3 $$ then $$ f(x) < 0 $$

*If $$x=0, x=1,x=3$$ then $$f(x)=0$$

*f   has a local maximum at 
$$
x_0  = \frac{1}
{5}\left( {8 - \sqrt {19} } \right)
$$
and
$$
f(x_0 ) = \frac{{2\left( {2147\sqrt {19}  - 8986} \right)}}
{{3125}} < \frac{1}
{2}
$$

*$$g(x)=-(x^2-3x+4) \leq -\frac{7}{4} \forall x \in \mathbb R$$
Therefore if
$$
0 \leqslant x \leqslant 3
$$
it is
$$
p(x) = f(x) + g\left( x \right) \leqslant \frac{1}
{2} - \frac{7}
{4} < 0
$$
It follows that if p(x)=0 then x>3
</s>