Proving inequality   $x^{10}-x^6+x^2-x+1>0$ How can the inequality  $x^{10}-x^6+x^2-x+1>0$ be proved
a) using elementary mathematical methods？
b) using higher mathematical methods？
 A: Elementary methods:
$x^{10}-x^6+x^2-x+1=0.5x^{10}+0.5(x^{10}-2x^6+x^2)+0.5(x^2-2x+1)+0.5=$
$=0.5x^{10}+0.5(x^5-x)^2+0.5(x-1)^2+0.5\ge 0.5>0$
There is fun fact: if we have polynomial $p(x)$, such that $p(x)\ge 0 \,\forall x\in R$, then there exists polynomials $g(x), h(x)$ such that:
$$
p(x)=(g(x))^2+(h(x))^2
$$
Though often it's not clear how to find such $g(x), h(x)$.
A: By completing square a few times you can get:
$x^{10}-x^6+x^2-x+1=$
$x^2(x^8-x^4)+x^2-x+1=$
$x^2\left(x^4-\frac12\right)^2+\frac34x^2-x+1=$
$x^2\left(x^4-\frac12\right)^2+\frac34\left(x-\frac23\right)^2+2/3$
This is a sum of positive expressions.
You can check this computation at wolframalpha:
http://www.wolframalpha.com/input/?i=x%5E2%28x%5E4-1%2F2%29%5E2%2B3%2F4%28x-2%2F3%29%5E2%2B2%2F3
A: Another elementary method:
We want to show that $x^{10}+x^2+1 >x^6+x$.  If we can show it for $x \ge 0$, it will be true always.  This is because if $x$ is positive, replacing $x$ by $-x$ does not change $x^{10}+x^2+1$, but turns  $x^6+x$ into the smaller $x^6-x$.
We now take care of  $x \ge 0$.  If $x\ge 1$, then $x^{10}\ge x^6$ and $x^2 \ge x$, so 
$x^{10}+x^2+1 >x^6+x$.
If $0 \le x < 1$, then $x^2 \ge x^6$ and $1 >x$, so $x^{10}+x^2+1 >x^6+x$.
A: Elementary, Comparative Method
First, let's rewrite this as $x^{10}$ +$x^2$< $x^6$+$x$-$1$.
Let's take it case by case.  Let x be negative and less than one.  We can immediately see that the left side of the inequality is going to be positive.  Since x is negative, x-1 is going to be less than $x^2$.
  Now, we can compare $x^{10}$ to $x^6$ and immediately see that the magnitude of $x^{10}$ is greater.  Thus, for this case, the inequality holds.
The same argument holds if x is greater than 1, as x-1 will still be less than $x^2$, and $x^{10}$ will be greater than $x^6$.  Similar arguments work for when the magnitude of x is less than 1.
