Question: Find the sum of all integral values of '$a$' such that $$a(x^2+x-1) \leq(x^2+x+1)^2 \text { for all } x \in R$$
My attempt:
I thought about representing $$a\leq \frac{(x^2+x+1)^2}{(x^2+x-1)}$$
but as $x\rightarrow \infty$, This goes to $\infty$ also. I don't really know the approach further. Please help out. I considered quadratic equations for a while but no fruitful result.
Hint: write your inequality as $$a(x^2+x+1-2)\le (x^2+x+1)^2$$ and set $$t=x^2+x+1$$ then you have to solve $$a(t-2)\le t^2$$ can you proceed?