direct proof - prove the inequality

Question

[I have proved a similar inequality; here.][1]

I am trying to prove the inequality:

$$x + \frac 1x \le -2, x < 0$$

I can proceed like I did in [1].

Scratch work; like so:

$$x + \frac 1x \le -2$$

$$ x(x + \frac 1x) \le -2x$$

$$ x^2 + 1 \le 2x$$

$$ x^2 + 2x + 1 \le 0$$

$$ (x +1)^2 \le 0$$

To prove this, I have to derive the original expression from the expression:

$$ (x +1)^2 \le 0$$

But I don't know if I can because I do not think this is true.

If it were true I would prove the inequality by reversing the process and checking the algebra is correct; like so:

$$ x^2 + 2x + 1 \le 0$$ $$ x^2 + 1 \le -2x$$ $$x + \frac 1x \le -2$$ [1]: How to prove an inequality by mathematical induction?

I am pretty sure this is not correct because I do not think I obtained a true expression in my scratch work.

I am asking if you can verify this proof and identify errors; thanks.

Answer 1

It must be $$x^2+1\geq -2x$$ so we get $$(x+1)^2\geq 0$$ the inequality sign must be reverse when we multiplying by $x<0$

Answer 2

You made a couple errors that cancel each other out but are incorrect.

First of all, $(x+1)^2\color{red}\ge0$ whether $x>0, x=0, $ or $x<0$.

Second, if $x<0$, then reverse an inequality when multiplying or dividing both sides by $x.$

You should say as follows: $$x + \frac 1x \le -2, x < 0$$

$$ x(x + \frac 1x) \color{red}\ge -2x, x < 0$$

$$ x^2 + 1 \ge \color{red}-2x$$

$$ x^2 + 2x + 1 \ge 0$$

$$ (x +1)^2 \ge 0$$

And you can write that upside down too.