We have $2(x^2+y^2+z^2) \geq \frac{2}{3}(x+y+z)^2=\frac{2}{9}(x+y+z)(3(x+y+z))$ and $xyz=x+y+z \geq 3\sqrt[3]{xyz}$ so $xyz \geq 3\sqrt{3}$ and $x+y+z \geq 3\sqrt[3]{xyz} \geq 3\sqrt[3]{3\sqrt{3}}=3\sqrt{3}$ so $\frac{2}{9}(x+y+z) \geq \frac{2\sqrt{3}}{3}>1$

my solution= by AM-GM $xyz\ge 3 (xyz)^{1/3}$ or $xyz\ge 3\sqrt 3$ now $2(x^2+y^2+z^2)\ge \frac{2(x+y+z)(x+y+z)}{3}\ge 2\sqrt{3} (x+y+z) > 3(x+y+z)$ so we are done

$x+y+z=xyz \leq (\frac{x+y+z}{3})^3 \implies x+y+z \geq 3\sqrt{3}$

sqing wrote: $x+y+z=xyz \leq (\frac{x+y+z}{3})^3 \implies x+y+z \geq 3\sqrt{3}$

Let $x$, $y$, $z$ be non-negative reals and $p=x+y+z$, $q=xy+yz+zc$, $r=xyz$ where $p=r$. We want to prove that \[ 2(p^2-2q) \ge 3p \]Squaring and homogenizing for $x,y,z$ yields the equivalent \[ (4p^4-8p^2q +8q^2)p \ge 9p^2r \]When $p=0$, we have equality. Otherwise, dividing both sides by $p$ and expanding it we have another equivalent \[ 4 x^4+8 x^3 y+8 x^3 z+16 x^2 y^2+15 x^2 y z+16 x^2 z^2+8 x y^3+15 x y^2 z+15 x y z^2+8 x z^3+4 y^4+8 y^3 z+16 y^2 z^2+8 y z^3+4 z^4 \ge 0 \]which is clearly true. Equality holds when $x=y=z=0$.

Nice, solution @seoneo, but i think it is much more complicated than the other proposed!

Pretty easy for a JBMOSL A5 right?

If one of $x,y,z$ is $0$, W.L.O.G. $x=0$, we have $y+z=0\implies y=z=0$, which is clear. If none of $x,y,z$ is $0$, let $x=1/a,b=1/y,$ and $c=1/z$ where $a,b,c\in \mathbb{R}^+$. We have $ab+bc+ca=1$. So, we want to prove that \[2\sum \frac{1}{a^2} \ge 3\sum \frac{1}{a} \iff \frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b} \ge \frac{3}{2}.\]By Cauchy Schwarz Engel, we have \[\sum \frac{ab}{c} =\sum \frac{(ab)^2}{abc}\ge \frac{1}{3abc}.\]By AM-GM, we have \[\frac{ab+bc+ca}{3}=\frac{1}{3} \ge \sqrt[3]{(abc)^2} \iff \sqrt{3}\le \frac{1}{3abc}\implies \sum \frac{ab}{c} \ge\frac{1}{3abc} \ge \sqrt{3} > \frac{3}{2}.\]Equality occurs when $x=y=z=0$.

Let $p=x+y+z$,$q=xy+yx+zx$,$r=xyz$ where $p=r$ the inequality becomes $$2p^2 -4q \geq 3p \leftrightarrow$$$$2p^2-4q-3p \geq 0$$ Fix $p=p_0$.From the condition r is fixed too, hence we have $$f(q)=2p^2-4q-3p \geq0$$ Note that $p^2 \geq 3q$ (AM-GM) substuting this we get $$f(q) \geq 2q-3p\geq0$$ Note that $2q-3p$ is linear to q hence this expression gets its minimal value at minimal value of q, by PQR Lemma q is at minimum value at x=y. Substuting x=y to the condition we get $x=y$ and $z=\frac{2x}{x^2-1}$ Hence $$p=r=\frac{2x^3}{x^2-1}$$, $$q=\frac{x^4+3x^2}{x^2-1}$$ using these at 2q-3p $$2 \cdot \frac{x^4+3x^2}{x^2-1}-3 \cdot \frac{2x^3}{x^2-1}=\frac{2x^4+6x^2-6x^3}{x^2-1}$$ Now we have to find minimum value of this expression, note that $$\frac{d \left( \frac{2x^4+6x^2-6x^3}{x^2-1} \right) }{dx}=\frac{2x(2x^4-3x^3-4x^2+9x-6)}{(x^2-1)^2}$$ note that this derivative gets to zero at $x=0,\sqrt{3},-\sqrt{3}$, x is positive integer hence we have to test $x=0$ and $x=\sqrt{3}$ After substuting we can see these holds the inequality and the equality case is at x=y=z=0 Is this right

Let $x,y$ and $z$ be non-negative real numbers satisfying the equation $x+y+z=xyz$. Prove that $$x^3+y^3+z^3\geq \sqrt 3 (x^2+y^2+z^2)$$

sqing wrote: Let $x,y$ and $z$ be non-negative real numbers satisfying the equation $x+y+z=xyz$. Prove that $$x^3+y^3+z^3\geq \sqrt 3 (x^2+y^2+z^2)$$ AM-GM: $$x^3+x^3+3\sqrt{3} \geq 3\sqrt{3}x^2$$Therefore: $$\sum{x^3}\ge\frac{1}{2} \sum{(3\sqrt{3}x^2-3\sqrt{3})}=\frac{3\sqrt{3}}{2}\sum{x^2}-\frac{9\sqrt{3}}{2}$$AM-GM again: $$x+y+z=xyz\le{(\frac{x+y+z}{3})^3} \Rightarrow x+y+z\ge3\sqrt{3}$$Then: $$3\sum{x^2}\ge(\sum{x})^2\ge27 \Rightarrow \sum{x^2}\ge9$$So: $$x^3+y^3+z^3\ge\frac{3\sqrt{3}}{2}\sum{x^2}-\frac{9\sqrt{3}}{2}\ge\sqrt{3}\sum{x^2}+\frac{\sqrt{3}}{2}\sum{x^2}-\frac{9\sqrt{3}}{2}\ge\sqrt{3}(x^2+y^2+z^2)$$