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$x,y,z \in \mathbb R$

My attempt:

Notice that if $x^2 + y^2 + z^2 \geq xyz$ and $x+y+z \geq xyz$. So : $$x^2 + y^2 + z^2 \geq x+y+z $$ But this is actually not always true, take the case when $ 0<x,y,z <1$ And you will see that this is not working.

Maybe I’m wrong.

PNT
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2 Answers2

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I guess you guys are too fast and / or assume too much prior knowledge. Let's digest it.

We need to be careful since all numbers are reals. Hence AM-GM and square root methods don't work.

First, note that if $xyz < 0$, then there is nothing to prove, since in this case $x^2 + y^2 + z^2 \geq 0 \geq xyz$ holds in all cases. So we need to consider $xyz \ge 0$ only.

We want to establish the following train of inequalities: $$ (x^2+y^2+z^2)^2 \ge (xy+yz+zx)^2 \ge {3xyz(x+y+z)} \ge {3} (xyz)^2 = 3 |xyz|^2 $$ The first one follows from $$ x^2+y^2+z^2 - ( xy+yz+zx) = \frac12 \Big[ (x-y)^2+(y-z)^2+ (z-x)^2 \Big] \ge 0 $$ The second one follows from $$ (xy+yz+zx)^2 - 3xyz(x+y+z) = \frac12 \Big[(xy-yz)^2+(zx-xy)^2+(yz-zx)^2 \Big] \ge 0 $$ For the third one, for $xyz > 0$, then this follows from the question's condition: $x + y+ z \ge xyz$.

So indeed it holds for all reals: $$ x^2+y^2+z^2 \ge \sqrt{3} |xyz| \ge \sqrt{3} \cdot xyz $$

This is also the tightest inequality, since we observe that equality holds for the question's condition and for this solution at $x=y=z = \sqrt{3}$. (River Li asked for that constant $k = \sqrt{3}$).

Andreas
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First, we prove for all positive $x,y,z$.

If all of $x,y,z\geq1$ we have
$$x^2\geq x ,\\ y^2\geq y \\ z^2\geq z$$ $$⇒x^2+y^2+z^2\geq x+y+z \geq xyz\\$$

If at least one of $x,y,z$ is less than $1$ then take (WLOG) $x\geq y\geq z>0$ & $1>z>0$

$$\frac{x}{y}\geq1 $$

$$⇒\frac{x}{yz}\geq1 $$ $$⇒\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}\geq 1$$ $$⇒x^2+y^2+z^2\geq xyz$$

If two of $x,y,z$ is negative (say $x$ and $y$) we have $$-x-y+z ≥ x+y+z ≥xyz=(-x)(-y)z$$ $$⇒-x-y+z ≥(-x)(-y)z$$ Here $-x,-y,z$ are positive numbers. So from the earlier result we can say, $$(-x)^2+(-y)^2+z^2\geq (-x)(-y)z$$ $$⇒x^2+y^2+z^2\geq xyz$$

The inequality is obvious when one of $x,y,z$ is $0$ or if one of $x,y,z$ is negative or if all of $x,y,z$ is negative.

Asher2211
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  • you said either $\frac{x}{y}\geq1 $ or $\frac{y}{x}\geq1 $ is true; then why did you use both inequalities in :$$\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}\geq 1$$ – PNT Feb 10 '21 at 19:09
  • @yassir I realised it now. No need to use both inequality – Asher2211 Feb 10 '21 at 20:08
  • You forgot the case when two of $x,y,z$ are negative and one is positive. – Andreas Feb 10 '21 at 20:15
  • Yeah you’re right, i think that @Asher Ambraham’s method won’t works in this case – PNT Feb 10 '21 at 21:21
  • @Andreas I have included that case in the post now. – Asher2211 Feb 11 '21 at 06:36
  • $$(-x)^2+(-y)^2+z^2\geq (-x)(-y)z$$ this inequality works only if $x,y,x \geq 1$?? – PNT Feb 11 '21 at 14:37
  • @yassir I proved $x^2+y^2+z^2≥xyz$ if $x,y,z$ is positive(not just for $x,y,z \geq 1$.) Therefore $(-x)^2+(-y)^2+z^2\geq (-x)(-y)z$ if x and y are negative. – Asher2211 Feb 11 '21 at 15:18