The inequality $x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+3/4 >0$ holds for all $x\in\mathbb R$ 
Show
$\forall \ x \in \mathbb{R}:\quad  x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+\dfrac{3}{4}>0$ 

My attemps:
Case $x=-1$ 
That is true for this case
Then for $x \neq - 1$:
$$\dfrac 3 4 - x + x^2 - x^3 + x^4 - x^5 + x^6 = \dfrac{1 + x^7}{1 + x} - \dfrac 1 4$$
let  $g(x)=\dfrac{1 + x^{7}}{1 + x}-\dfrac{1}{4} \quad \forall x\in \mathbb{R} \backslash  \{-1\}$
then
$
\begin{align*}
g'(x)&=\dfrac {(1+x^7)'(1+x)-(1+x)'(1+x^7)}{(1 + x)^2}  \quad \forall x\in \mathbb{R} \backslash  \{-1\}\\
g'(x)&=\dfrac {(7 x^6)(1+x)-(1+x^7)}{(1 + x)^2}  \quad \forall x\in \mathbb{R} \backslash  \{-1\}\\
g'(x)&=\dfrac {7 x^6+7 x^7-1-x^7)}{(1 + x)^2}  \quad \forall x\in \mathbb{R} \backslash  \{-1\}\\
g'(x)&=\dfrac{7 x^6+6 x^7-1}{(1 + x)^2}  \quad \forall x\in \mathbb{R} \backslash  \{-1\}
\end{align*}
$
To determine the sign of the numerator $(7 x^6+6 x^7-1)$
once time let  :$ h(x)=7 x^6+6 x^7-1$
then
$$h'(x)=42x^5+42x^6=42(1+x)x^5$$
$$h'(x)=0  \Longleftrightarrow x=-1 \text{or} x=0$$
thus $h$ admits a minimum on the point $x=0$
and a maximum on the point $x=-1$
or $h(-1)=0$ and $h(0)=-1$
$\lim_{x\to -\infty}h(x)=-\infty$ and $\lim_{x\to +\infty}h(x)=+\infty$
as $h(-1)=0$ and $h(0)=-1$ by Intermediate value theorem  there is $u \in(-1, 0)$ such that h(u) = 0.
i'm stuck here


*

*am i on my way ?

*is there any other ways to solve it

 A: You can write your expression as
$$x(x-1)(x^4+x^2+1)+\frac{3}{4}\ .$$
Now consider $x(x-1)(x^4+x^2+1)$: the third factor is always positive, hence the whole assume negative values only for $x\in(0,1)$. In this interval $x(x-1)$ is negative and $x^4+x^2+1<3$. The minimum of $x(x-1)$ in $(0,1)$ is $-\frac{1}{4}$ for $x=\frac{1}{2}$ hence you have 
$$x(x-1)(x^4+x^2+1)>-\frac{3}{4}$$
For $x\not\in(0,1)$  $x(x-1)(x^4+x^2+1)\geq 0$ so it's ok.
A: Another way:  Sum the obvious AM-GMs:
$$\tfrac12 x^6 + \tfrac12 x^4 \ge x^5, \quad \tfrac12 x^4 + \tfrac12 x^2 \ge x^3$$
$$\tfrac12 x^2 + \tfrac34 \ge \sqrt{\tfrac32}|x| \ge x \implies \tfrac12 x^2 + \tfrac34 > x$$
with $\frac12x^6 \ge 0$.
A: Another way would be to prove $(1 + x^7) - {1 \over 4}(1 + x) > 0$ for $x > -1$, and $(1 + x^7) - {1 \over 4}(1 + x) <  0$ for $x < -1$, using calculus on $f(x) = (1 + x^7) - {1 \over 4}(1 + x) = x^7 - {1 \over 4}x + {3 \over 4}$. Since $\lim_{x \rightarrow \infty} f(x) = \infty$ and $\lim_{x \rightarrow -\infty} f(x) = -\infty$, we need show the minimum of $f(x)$ for $x > -1$ is positive and the maximum of $f(x)$ for $x < -1$ is negative.
One has $f'(x) = 7x^6 - {1 \over 4}$, which is equal to zero at $x = \pm (28)^{-{1 \over 6}}$, both of which are greater than $-1$. Hence $f$ is increasing for $x > -1$, so since $f(-1) = 0$, this gives the $x < -1$ part.
As for $x > -1$, note that $f''(x) = 28x^5$, which is positive at $x =   (28)^{-{1 \over 6}}$ and negative at $-(28)^{-{1 \over 6}}$. So $f$ has a local minimum at $x =   (28)^{-{1 \over 6}}$ and a local maximum at $x = -(28)^{-{1 \over 6}}$. So we just have to plug in the values $x = -1$ (the left endpoint) and $x = (28)^{-{1 \over 6}}$ (the minimum). We saw $f(-1) = 0$, and plugging in $x = (28)^{-{1 \over 6}}$ gives $(28)^{-{7 \over 6}} - {1 \over 4} (28)^{-{1 \over 6}} + {3 \over 4} = .62..$ which is positive. So $f(x)$ is positive for $x > -1$.
