2
$\begingroup$

How to prove the following inequality by mathematical induction?

$$x + \frac 1x \ge 2, x \gt 0$$

I am aware of this.

First, I have to prove $P(1)$; then $P(n+1)$.

I am stuck at $P(n+1)$ because I do not know how to add the plus one to it. thanks.

EDIT 4/10/19

I misunderstood the problem; I thought I needed to prove the above inequality by mathematical induction; I learnt that I just need to prove it. I think the easiest way to prove this inequality is by constructing a direct proof.

$\endgroup$

put on hold as unclear what you're asking by José Carlos Santos, Lord Shark the Unknown, ancientmathematician, Paras Khosla, Cesareo yesterday

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ And what is $P(n)$? $\endgroup$ – José Carlos Santos Mar 19 at 23:09
  • $\begingroup$ Induction on what? $\endgroup$ – Bernard Mar 19 at 23:10
  • $\begingroup$ Is $x$ here just positive integers, or all positive real numbers? $\endgroup$ – Minus One-Twelfth Mar 19 at 23:14
1
$\begingroup$

No way! Induction is only for integers.

By the way here’s another way to solve it:

$$(\sqrt{x}-\frac{1}{\sqrt{x}})^2\ge 0$$ $$\Rightarrow x +\frac{1}{x}-2\ge 0\Rightarrow x+\frac{1}{x}\ge 2$$

$\endgroup$
1
$\begingroup$

If the inequality is supposed to be proved for positive integers $x$ then you get $P(n+1)$ from $P(n)$ as follows: $n+1+\frac 1 {n+1}=n+\frac 1 n +1+\frac 1 {n+1}-\frac 1 n \geq 2+1-\frac 1 {n(n+1)}\geq 2$ because $n(n+1) \geq 1$ and $\frac 1 {n(n+1)} \leq 1$

$\endgroup$
  • $\begingroup$ what if $n=1$?? then $1(1+1) \geq 1 $ may not be true $\endgroup$ – James Mar 20 at 0:06
  • $\begingroup$ @JimmySabater I don't see any problem here when $n=1$. $\endgroup$ – Kavi Rama Murthy Mar 20 at 0:10
  • $\begingroup$ What do you even mean by saying $2 \geq 1$ may not be true? $\endgroup$ – Kavi Rama Murthy 2 days ago
1
$\begingroup$

I think the easiest way to prove this inequality is by direct proof where you simplify the inequality, thereby obtaining a true statement. Then you try to check the algebra is still valid; like so:

$$x + \frac 1x \ge 2$$

$$ x(x + \frac 1x) \ge 2x$$

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

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

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

Once you simplify, you prove it by reversing the process and checking the algebra is correct; like so:

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

$\endgroup$
0
$\begingroup$

The result follows if one puts $a=x$ and $b=1/x$ on the famous AM-GM inequality

$\endgroup$
  • 3
    $\begingroup$ The question clearly says 'by mathematical induction'. $\endgroup$ – Kavi Rama Murthy Mar 20 at 0:10

Not the answer you're looking for? Browse other questions tagged or ask your own question.