Inequalities Proof [closed]

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

$$1/x + 1/y + 1/z ≥3?$$

Thanks!

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

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

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

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

Only if x, y and z are positive reals.

It's wrong.

Try $$z\rightarrow0^-$$.

But for positive variables by C-S we obtain: $$\sum_{cyc}\frac{1}{x}=\frac{1}{x+y+z}\cdot\sum_{cyc}x\sum_{cyc}\frac{1}{x}\geq\frac{1}{x+y+z}\cdot(1+1+1)^2\geq3.$$ This is @Blue's very nice visual proof from trigonography.com that

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