Think Again, Please

Idea: minimize the first $k$ terms of the sum, for each $k = 2, 3, 4, 5$ . This works because $x \mapsto x + \frac{1}{x}$ is increasing for $x > 1$ .

Let $g(x) = x + \frac{1}{x}$ . Then $f(x) = g(g(g(g(x^2))))$ , and we want to find the minimum value of $f$ over all $x \neq 0$ (when $x = 0$ , $g(x)$ is not defined; we'll see that $f$ is defined for all other $x$ ).

Observe that if $x > 1$ , then $g(x)$ is an increasing function. This is easy to see: if $1 < x < y$ , then

$\begin{aligned} g(y) - g(x) &= \left( y + \frac{1}{y} \right) - \left( x + \frac{1}{x} \right) \\ &= (y-x) + \left( \frac{1}{y} - \frac{1}{x} \right) \\ &= (y-x) + \frac{x-y}{xy} \\ &= (y-x) \left( 1 - \frac{1}{xy} \right) \end{aligned}$

Since $y > x$ , $y-x > 0$ . Since $x,y > 1$ , then $xy > 1$ , so $1 - \frac{1}{xy} > 0$ . So $g(y) - g(x) > 0$ , proving that $g$ is increasing for $x > 1$ .

Then, note that for $x > 0$ , $g(x) \ge 2$ . This uses AM-GM inequality; if $x > 0$ , then $\frac{1}{x} > 0$ , so $x + \frac{1}{x} \ge 2 \sqrt{x \cdot \frac{1}{x}} = 2$ . The important point is that $g(x) \ge 2 > 1$ .

Now, to the problem. Since $x^2 > 0$ , we know $g(x^2) \ge 2$ , so we know $g(g(x^2)) \ge 2$ and $g(g(g(x^2))) \ge 2$ . We showed that $g(x)$ is increasing for $x > 1$ , so to minimize $f(x) = g(g(g(g(x^2))))$ , we need to minimize $g(g(g(x^2)))$ , which in order means we need to minimize $g(g(x^2))$ and thus $g(x^2)$ . Since we know the minimum value of $g(x^2)$ is 2 (achieved when $x = 1$ or $x = -1$ ), the minimum value of $g(g(x^2))$ is achieved when $g(x^2)$ is as small as possible, in this case 2. This means $g(g(x^2)) \ge 2 + \frac{1}{2} = \frac{5}{2}$ . Continuing in the same manner, $g(g(g(x^2))) \ge \frac{5}{2} + \frac{2}{5} = \frac{29}{10}$ , and finally $f(x) = g(g(g(g(x^2)))) \ge \frac{29}{10} + \frac{10}{29} = \boxed{\frac{941}{290}}$ .

Note: This is a shortcut approach and may not necessarily work in other cases. For the correct solution, see Ivan Koswara's solution to this problem.

For all real $x$ we know that $x^2$ must be positive, therefore by applying AM-GM inequality we have

$\dfrac 12 \cdot \left( x^2 + \dfrac{1}{x^2} \right) \ge \sqrt{x^2 \cdot \dfrac{1}{x^2}} \implies \color{#3D99F6}{x^2 + \dfrac{1}{x^2} \ge 2}$

And so the minimum of $x^2 + \dfrac{1}{x^2}$ is 2.

It can be easily seen that

$\begin{aligned} \min \{f(x)\} &= \underbrace{x^2 + \dfrac{1}{x^2}}_{\color{#3D99F6}{2}} + \dfrac{1}{\underbrace{x^2 + \dfrac{1}{x^2}}_{\color{#3D99F6}{2}} } + \dfrac{1}{\underbrace{x^2 + \dfrac{1}{x^2}}_{\color{#3D99F6}{2}} + \dfrac{1}{\underbrace{x^2 + \dfrac{1}{x^2}}_{\color{#3D99F6}{2}}}} +\dfrac{1}{\underbrace{x^2 + \dfrac{1}{x^2}}_{\color{#3D99F6}{2}} + \dfrac{1}{\underbrace{x^2 + \dfrac{1}{x^2}}_{\color{#3D99F6}{2}}} + \dfrac{1}{\underbrace{x^2 + \dfrac{1}{x^2}}_{\color{#3D99F6}{2}} + \dfrac{1}{\underbrace{x^2 + \dfrac{1}{x^2}}_{\color{#3D99F6}{2}}}}} \\ &= 2 + \dfrac 12 + \dfrac{1}{2+\dfrac 12} + \dfrac{1}{2+\dfrac 12 + \dfrac{1}{2+\dfrac 12}} \\ &= \dfrac 52 + \dfrac 25 + \dfrac{1}{\dfrac 52 + \dfrac 25} \\ \\ &= \dfrac{29}{10} + \dfrac{10}{29} \\ \\ &= \boxed{\dfrac{941}{290}} \end{aligned}$

Consider last term and other terms except last term

Then it will be in the form of t + $\frac{1}{t}$ .

To be min t should be min (since the function is increasing function because t>2)

So now to min t again take t's last term and other terms except last term of t's.

Again same procedure

Continue like this and will end up with minimising u + 1/u

Where u = $x^2$ + $x^-2$

So minimising (u + 1/u) is to minimize u

So it will be minimum at x = 1

So you will get $f(x)$ ≥ $\frac{29}{10}$ + $\frac{10}{29}$ = $\frac{941}{290}$

Let $g\left(x\right)=x^{2}+\frac{1}{x^{2}}-\frac{1}{x^{2}+\frac{1}{x^{2}}}-\frac{1}{x^{2}+\frac{1}{x^{2}}+\frac{1}{x^{2}+\frac{1}{x^{2}}}}-\frac{1}{x^{2}+\frac{1}{x^{2}}+\frac{1}{x^{2}+\frac{1}{x^{2}}}+\frac{1}{x^{2}+\frac{1}{x^{2}}+\frac{1}{x^{2}+\frac{1}{x^{2}}}}}\geq\frac{219}{290}$ (AM-GM for each part) $\Rightarrow ming\left(x\right)=g\left(1\right)=\frac{219}{290}$

We have $f\left(x\right)+g\left(x\right)=2\left(x^{2}+\frac{1}{x^{2}}\right)=h\left(x\right)\geq2.2=4$ (AM-GM) $(minh\left(x\right)=h\left(1\right)=4)$

Moreover $h\left(x\right)$ attain minium value when $x=1$ and $g\left(x\right)$ also attain minium value when $x=1$ , so $f\left(x\right)$ must attain minium when $x=1$ (all of functions are increasing function)

So we have $minf\left(x\right)=f\left(1\right)=h\left(1\right)-g\left(1\right)=\frac{941}{290}$ .