Can the minimum value of x + 1/x ever be less than 1 on x > 0?

[Rectifier]

The problem
I want to solve the following inequality:
$$ x+\frac{1}{x}<1 $$

The attempt
$x+\frac{1}{x}<1 \\ x+\frac{1}{x}-1<0 \\ \frac{x^2}{x}+\frac{1}{x}-\frac{x}{x}<0 \\ \frac{x^2-x+1}{x}<0$ $x \neq 0$

I tried to factor the numerator to examine the polynomial with a character table but it has complex roots.

 

[mfb]

I tried to factor the numerator to examine the polynomial with a character table but it has complex roots.

This tells you the numerator never gets negative.
There are solutions, you still have the denominator there to consider.

 

[Rectifier]

This tells you the numerator never gets negative.

Ehm, why is that? I am not sure how to prove that it can't get negative.

 

[mfb]

It does not have real roots, so it cannot be zero anywhere. It has to be either always positive or always negative, and it is easy to check which case you have.

 

[Rectifier]

Ehm, why is that? I am not sure how to prove that it can't get negative.

I am answering my own question :D

I guess that I should plot the function and see that it is a parabola that hovers over the x-axis. Which means that it is >0 for all x. Thus + everywhere in the character table.

 

[Rectifier]

Thank you for the help, mfb!

 

[SammyS]

It does not have real roots, so it cannot be zero anywhere. It has to be either always positive or always negative, and it is easy to check which case you have.

Thank you for the help, mfb!

What's your final solution for the initial inequality?

$\displaystyle x+\frac{1}{x}<1 \$

 

[Rectifier]

What's your final solution for the initial inequality?

$\displaystyle x+\frac{1}{x}<1 \$

It is that the inequality is true for $x<0$.

 

[mfb]

Right.

It is probably easier to show that if you consider the three cases x<0, 0<x<1 and x>=1 separately, but your approach works as well.

 

[Ray Vickson]

Right.

It is probably easier to show that if you consider the three cases x<0, 0<x<1 and x>=1 separately, but your approach works as well.

Of course, another way is to look at the minimum of $f(x) = x + 1/x$ in $x > 0$, to see if $f(x)$ can ever be < 1 on $x > 0$.

One way to find the minimum is to use calculus, but in the spirit of a pre-calculus argument we can do it another way. We have $f(x) = (a+b)/2$, where $a = 2 x, b = 2/x$ The arithmetic-geometric inequality says that for positive terms $a,b$ we have $(a+b)/2 \geq \sqrt{ab}$, with equality holding if and only if $a = b$. Thus, $f(x) \geq \sqrt{2x \cdot 2/x} =2$ for all $x > 0$, and equality is obtained when $x = 1/x$. In other words, 2 is the minimum value of $f(x)$ on $x > 0$, and occurs at $x = 1$.

This argument is not particularly clever; it just applies basic "Geometric Programming" (GP) ideas to the current, simple case. See, eg.,
http://www.mpri.lsu.edu/textbook/chapter3.htm
for a gentle introduction to GP..