Q. Show that the function $f(x)$ defined below attains a unique minimum for $x 0$. What is the minimum value of the function? What is the value of $x$ at which the minimum is attained?
$$\hspace{3cm} f(x) = x^2+x+\frac1x+\frac1{x^2}\quad \text{for}\;\;x\neq0$$
Is there an elegant way of proving this? Instead of differentiating twice and solving cubic equation.

Question

Q. Show that the function $f(x)$ defined below attains a unique minimum for $x > 0$. What is the minimum value of the function? What is the value of $x$ at which the minimum is attained?
$$\hspace{3cm} f(x) = x^2+x+\frac1x+\frac1{x^2}\quad \text{for}\;\;x\neq0$$
Is there an elegant way of proving this? Instead of differentiating twice and solving cubic equation.

anonymous · Answer

this looks simmilar to Laurent series

blockcolder · Answer

Rewrite the function like this:
$$f(x)=x^2-2+\frac{1}{x^2}+x-2+\frac{1}{x}+4\
=\left( x-\frac{1}{x}\right)^2+\left( \sqrt{x}-\frac{1}{\sqrt{x}}\right)^2+4.$$

anonymous · Answer

Is proving that some function $g(x)$ defined as $g(x) = x +\frac{1}{x}\quad$for$\;x>0$, is minimized when $x=1$, sufficient to prove $h(x) = u(x)+\frac{1}{u(x)}\quad$for $x>0$ is minimized when $u(x)=1$?

blockcolder · Answer

You don't need to see it that way. Look at it this way: The expression is composed of squares, which attain a minimum value of 0 at x=1.

anonymous · Answer

Ohh hmm sorry :/
Yeah, that's much better.

anonymous · Answer

Nice blockcoder. Thanks.

I wonder why it didn't cross my mind :/