I believe there will be values of $x$ for which the inequality $x^3 – 2x + 2 \ge 3 – x^2$ is true and values for which it is not true, because:
- LHS asymptotically increases but RHS decreases for increasingly positive values of $x$
- LHS asymptotically decreases faster than RHS for increasingly negative values of $x$
But I don't know how to reason much further about $x^3 – 2x + 2 \ge 3 – x^2$.
My idea is that if I get the inequality in a form I can reason about, I can determine what values of $x$ will satisfy the condition. I can reason about
$$\frac{a}{b} > 0$$
since either $a,b > 0$ or $a,b < 0$ for $\frac{a}{b} > 0$.
I can also reason about the signs of $a,b$ if they are expressed as a product of factors, since any even multiple of negative factors gives a positive (or zero) product. I think this will put me in a good spot to reason about what conditions must be met for the signs to satisfy the inequality $\frac{a}{b} > 0$ (although in my case, I am including zero).
Thus, my plan is to transform the inequality into the form LHS = a quotient of factorised terms and RHS = 0. But I'm not sure how.
Best Answer
$$x^3+x^2-2x-1 \ge 0$$
First, let's solve the equation: $$x^3+x^2-2x-1=0$$ Let's say $x=y-\frac{1}{3}$ : $$y^3-\frac{7}{3}y-\frac{7}{27}=0$$ We know that the: $(m+n)^3-3mn(m+n)-(m^3+n^3)=0$. Based on this, let's say $y=m+n$. Now let's calculate the following system of equations: $$\begin{cases} 3mn=\frac{7}{3} \\ m^3+n^3=\frac{7}{27} \end{cases}$$ To solve this system of equations, it is necessary to solve the equation below: $$m^3+\left(\frac{7}{9m}\right)^3=\frac{7}{27}$$ $$\frac{729m^6-189m^3+343}{729m^3}=0$$ Substitute $t$ for $m^3$ $$729t^2-189t+343=0$$ From here: $$t=\frac{7\sqrt{3} i}{18}+\frac{7}{54}$$ $$t=-\frac{7\sqrt{3} i}{18}+\frac{7}{54}$$ Since $m=t^3$ ,the solutions are obtained by solving the equation for each $t$ and variable $m \neq 0$:
$$m=\frac{{\sqrt{7}e}^\frac{-\arctan(3\sqrt{3})i+2πi}{3}}{3}$$ $$m=\frac{{\sqrt{7}e}^\frac{-\arctan(3\sqrt{3})i+4πi}{3}}{3}$$ $$m=\frac{{\sqrt{7}e}^\frac{\arctan(3\sqrt{3})i}{3}}{3}$$ $$m=\frac{{\sqrt{7}e}^\frac{-\arctan(3\sqrt{3})i}{3}}{3}$$ $$m=\frac{{\sqrt{7}e}^\frac{\arctan(3\sqrt{3})i+2πi}{3}}{3}$$ $$m=\frac{{\sqrt{7}e}^\frac{\arctan(3\sqrt{3})i+4πi}{3}}{3}$$ We can find $n$ using $m$, $y$ using $m$ and $n$, and $x$ using $y$:
$$x_1=\frac{ -\sqrt{7}cos\left( \frac{arccos\left( \frac{\sqrt{7}}{14} \right) }{3} \right) -\sqrt{21}sin\left( \frac{arccos\left( \frac{\sqrt{7}}{14} \right) }{3} \right)-1}{3} \approx −1.801937736$$
$$x_2=\frac{ 2\sqrt{7}cos\left( \frac{arccos\left( \frac{\sqrt{7}}{14} \right) }{3} \right) -1}{3} \approx 1.246979604$$
$$x_3=\frac{ \sqrt{21}sin\left( \frac{arccos\left( \frac{\sqrt{7}}{14} \right) }{3} \right) -\sqrt{7}cos\left( \frac{arccos\left( \frac{\sqrt{7}}{14} \right) }{3} \right)-1}{3} \approx −0.445041868$$ So we have to solve the following inequality: $$(x-x_1)(x-x_2)(x-x_3) \ge 0$$
$$(x+1,8)(x-1,25)(x+0,45)\ge 0$$ $$x \in \left[-\frac{9}{5},-\frac{9}{20}\right] \cup \left[\frac{5}{4},+∞\right)$$