Inequalities Proof

Question

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

$$1/x + 1/y + 1/z ≥3?$$

Thanks!

Answer 1

Only if x, y and z are positive reals.

Answer 2

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

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

Answer 3

It's wrong.

Try $z\rightarrow0^-$.

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

Answer 4

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

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