Find the maximum of positive integer $k$ so that for all positive real numbers $x$ we have: $x^6+x^5+x^4+x^2+x+4>kx^3$ 
Find the maximum of positive integer $k$ so that for all positive real numbers $x$ we have: $x^6+x^5+x^4+x^2+x+4>kx^3$

Since the power of the polynomial on LHS is greater than 3 I have no idea for it!

Assume $k \ge 9$. Then we have  $$9=1^6+1^5+1^4+1^2+1+4>k \times 1^3 \ge 9$$
Contradiction. So we have $k<9$. 
Note that $$x^6+x^5+x^4+x^3+x^2+x+4=(x^6+4)+(x^5+x)+(x^4+x^2)>8x^3$$
As follows from $\text{AM-GM}$, and because the equality condition can not be satisfied, there is no equality symbol.
So since $8$ is possible, but $k$ is the maximum, we have $k \ge 8$. 
So since $k$ is a positive integer that satisfies $$9>k \ge 8$$ we have $k=8$. The answer is $k=8$.  
