# how to solve the inequality?

The following two functions

$$f_1(x) = \frac{(3-x)(1-x)}{3}$$

and

$$f_2(y) = \frac{(1-y^2)}{(1+y) (2(1+y) - y(1-y)^2) }$$

are given.

Find range of values of $x$ and $y$ so that

$$f_1(x) < f_2(y)$$

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$x=1$, $y=0{}{}$ –  martini Nov 15 '12 at 13:12

We have $$2(1 + y) - y(1-y)^2 = 2 + 2y - y + 2y^2 - y^3 = -y^3 + 2y^2 + y + 2$$ and hence $$f_2(y) = \frac{1-y}{-y^3 + 2y^2 + y + 2}$$ Now let $y$ be given, we will find the $x \in \mathbb R$ with $f_1(x) < f_2(y)$, we have \begin{align*} f_1(x) &< f_2(y)\\ \iff x^2 - 4x + 3 &< 3f_2(y)\\ \iff (x-2)^2 - 1 &< 3f_2(y)\\ \iff (x-2)^2 &< 3f_2(y) + 1\\ \end{align*} We have \begin{align*} 3f_2(y) + 1 &= \frac{-y^3+2y^2-2y + 5}{-y^3 + 2y^2 + y + 2}\\ &= \frac{y^3 -2y^2 + 2y - 5}{y^3 - 2y^2 - y -2} \end{align*} As we want $(x-2)^2 < 3f_2(y) + 1$, we want this to be positive, Wolfram|Alpha tells us (I'm sure, one can do this by hand using Cardano's formulas to find the points where numerator and denominator change sign) that this is the case for $\def\ytwo{\frac13\left(2 + \sqrt[3]{44-3\sqrt{177}}+\sqrt[3]{44+3\sqrt{177}}\right)}\def\yone{\frac 13\left(2 - 2 \sqrt[3]{\frac 2{155 + 3\sqrt{1473}}}+\sqrt[3]{\frac12\left(155 + 3\sqrt{1473}\right)}\right)}$ $y \not\in \left[\yone, \ytwo\right]$ Then we can continue $$(x-2)^2 < 3f_2(y) + 1 \iff |x-2| < \sqrt{3f_2(y) + 1}$$ that gives $\def\ftwoy{\frac{y^3 -2y^2 + 2y - 5}{y^3 - 2y^2 - y -2}}$ $$x \in \left(2-\sqrt{\ftwoy}, 2+\sqrt{\ftwoy}\right).$$

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