I have two functions, $f(x)= x^2+x+1$ and $g(x)= \frac{1}{x}$. I have to prove that their point of intersection has a distance from the origin $<2$.
This is a high school problem so the proof shouldn't use more then derivatives.
Any hint is appreciated.
Firstly, lets find the point of intersection by equating the two functions:
$x^{2} + x+1 = \frac{1}{x}$
$x^{3} + x^{2} + x -1=0$
By considering the graph of this function, we see that it has $1$ real root. Calculating this root is possible, but not necessary. Instead we must find the interval in which this root exists by looking at sign changes.
Let $h(x) = x^{3} + x^{2} + x -1$
$h(1) > 0$, $h(\frac{1}{2})<0$
Lets call the real root $\alpha$, then we have $\frac{1}{2} < \alpha < 1$
Now the $y$ coordinate of the point of intersection is at $g(\alpha) = \frac{1}{\alpha}$
So the point of intersection is at $(\alpha, \frac{1}{\alpha})$
The distance, $D$, from the origin to this point is given by Pythagoras:
$D(\alpha) = \sqrt{\alpha ^{2} + \frac{1}{\alpha ^{2}}}$
$\frac{dD}{d\alpha} = 2\alpha - \frac{2}{\alpha ^{3}} = \frac{2(\alpha ^{4} - 1)}{\alpha ^{3}} < 0$ for $\frac{1}{2} < \alpha < 1$
Therefore $D$ is strictly decreasing in this domain.
Consider what value of $c$ would give $D(c) = 2$:
We find that $c$ = $\sqrt{2 - \sqrt{3}}$
Now we check the sign of $h(c)$:
$h(c) = (2-\sqrt{3})^{\frac{3}{2}} + 2 - \sqrt{3} + (2-\sqrt{3})^{\frac{1}{2}} -1$
$h(c) = (2-\sqrt{3})^{\frac{1}{2}}(3-\sqrt{3}) + 1-\sqrt{3}$
$h(c) = \sqrt{3}(\sqrt{3}-1)(2-\sqrt{3})^{\frac{1}{2}} + 1-\sqrt{3}$
$h(c) = (\sqrt{3} -1)(\sqrt{3}(2-\sqrt{3})^{\frac{1}{2}} -1)$
Now to prove that $h(c) < 0$, we must prove that:
$\sqrt{3}(2-\sqrt{3})^{\frac{1}{2}} < 1$
Since both sides are positive we can do the following:
$\Leftrightarrow 1 > 3(2-\sqrt{3}) \Leftrightarrow 3\sqrt{3} > 5 \Leftrightarrow 27 > 25$ which is true, hence the initial statement is also true since every step was an if and only if statement.
$\Rightarrow h(c) < 0$ which means $c$ is our new lower-bound for $\alpha$ $ \Rightarrow c<\alpha < 1$
Since $D$ is strictly decreasing, we have that the maximum value of $D$ would occur at the smallest value of $\alpha$(the value of $\alpha$ is of course fixed, we are just considering the limiting case given our constraints on $\alpha$). Since we found $c$ by setting $D(c) = 2$, and $c<\alpha < 1$, it follows that $D<2$.
You do not need to find the roots of a third degree polynomial to prove your inequality.
The given parabola and hyperbola intesect at a point whose abscissa $x\in(0,1)$. Its squared distance from the origin is given by $x^2+\frac{1}{x^2}$, that is a number greater than $2$. We have to prove it is less than $4$. That is the same as proving that $x^2> 2-\sqrt{3}$. Since $g(x)=x^2+x+1-\frac{1}{x}$ is an increasing function on $[0,\sqrt{2-\sqrt{3}}]$, it is enough to check that $g(\sqrt{2-\sqrt{3}})<0$.
Hint: the point of intersection satisfies $P(x)=x^{3} + x^{2} + x -1=0\,$, so for that point:
$$x^3 = -x^2-x+1 \tag{1}$$
The distance to the origin is $r = \sqrt{x^2+y^2}=\sqrt{x^2+\frac{1}{x^2}}\,$. Squaring, eliminating denominators, and using $(1)$ to eliminate powers of $x$ higher than $2$ gives:
$$ \require{cancel} \begin{align} r^2 x^2 & = x^4+1 \\ & = x(-x^2-x+1) + 1 \\ & = -x^3 -x^2 +x +1 \\ & = -(\cancel{-x^2}-x+\bcancel{1}) -\cancel{x^2} + x + \bcancel{1} \\ & = 2 x \end{align} $$
Therefore $r^2 = \cfrac{2}{x}\,$, so all that's left to prove is that $x \gt \cfrac{1}{2}\,$ which follows from $P\left(\cfrac{1}{2}\right)\cdot P(1) \lt 0$.
HINT Equate the two equations and then prove that the $x$ coordinate of the intersection point is in the interval $(\frac{1}{2}, 1)$. This should be enough find bounds of $y$ and then we can use the Distance Formula to prove that the distance is less than $2$.