0
$\begingroup$

if $x+y+z ≤ 3$ is it necessarily true that

$$1/x + 1/y + 1/z ≥3?$$

Thanks!

$\endgroup$

closed as off-topic by Thomas Shelby, Gibbs, Vinyl_cape_jawa, Song, mau Mar 13 at 15:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Thomas Shelby, Gibbs, Vinyl_cape_jawa, Song, mau
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ If $x,y,z$ can take negative real values then no it isn't necessary. $\endgroup$ – Yadati Kiran Mar 12 at 11:32
  • 1
    $\begingroup$ What conditions do you have? What have you tried? $\endgroup$ – Parcly Taxel Mar 12 at 11:33
5
$\begingroup$

If $x, y, z$ are positive, we have the well-known inequalities

$$x + \dfrac 1x \ge 2 \qquad y + \dfrac 1y \ge 2\qquad z + \dfrac 1z \ge 2$$

Adding them all up we get

$$x+y+z + \dfrac 1x + \dfrac 1y + \dfrac 1z \ge 6$$

Which yields

$$\dfrac 1x + \dfrac 1y + \dfrac 1z \ge 3$$

$\endgroup$
2
$\begingroup$

Only if x, y and z are positive reals.

$\endgroup$
1
$\begingroup$

It's wrong.

Try $z\rightarrow0^-$.

But for positive variables by C-S we obtain: $$\sum_{cyc}\frac{1}{x}=\frac{1}{x+y+z}\cdot\sum_{cyc}x\sum_{cyc}\frac{1}{x}\geq\frac{1}{x+y+z}\cdot(1+1+1)^2\geq3.$$

$\endgroup$
1
$\begingroup$

enter image description here

This is @Blue's very nice visual proof from trigonography.com that

$$x+\frac{1}{x}\;\geqslant\; 2$$

$\endgroup$

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