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

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

 [/INST] 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 $\frac12x^6 \ge 0$.
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