sqing wrote: Let $x,y,z>0 $ and $\sqrt{xyz}=xy+yz+zx$. Prove that$$x+y+z\leq \frac{1}{3}.$$ Using $(a+b+c)^2 \geq 3(ab+bc+ac)$, we get: \[ x+y+z \leq \frac{1}{3}(\sqrt{\frac{xy}{z}}+\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}})^2=\frac{1}{3}\frac{(xy+yz+xz)^2}{xyz}=\frac{1}{3}\]

Trivial by the inequality $(ab+bc+ca)^2 \geq 3abc(a+b+c)$.

sqing wrote: Let $x,y,z>0 $ and $\sqrt{xyz}=xy+yz+zx$. Prove that$$x+y+z\leq \frac{1}{3}.$$ $$xyz=(xy+yz+zx)^2=(xy)^2+(yz)^2+(zx)^2+2xyz(x+y+z)\geq \frac{1}{3}(xy+yz+zx)^2+2xyz(x+y+z)=\frac{1}{3}xyz+2xyz(x+y+z)$$$$xyz\geq \frac{1}{3}xyz+2xyz(x+y+z)\iff x+y+z\leq \frac{1}{3}.$$

IfΣx≤1/3 is wrong ThenΣx>1/3,so we can choose r make rΣx=1/3, r<1 ∏x=(Σx)² so∏(xr)>(Σ(xr•yr))² a=xr b=yr c=zr thenΣa=1/3 and∏a>(Σab)² But(Σab)²=Σa²b²+2Σa²bc ≥3Σa²bc=3(Σa)∏a=∏a It’s wrong,so Σx≤1/3

Let $x,y,z > 0,xy + yz + zx \leq3(2\sqrt 3-3) .$ Prove that$$\sqrt{\frac{xy + yz + zx}{3} }+ 1\leq \sqrt[3]{ (x + 1)(y + 1)(z + 1)}$$Let $x,y,z > 0,xy + yz + zx >3(2\sqrt 3-3) .$ Prove that$$\sqrt{xy + yz + zx }+ 1<\sqrt{ (x + 1)(y + 1)(z + 1)}$$(Florentin Visescu)

$\sqrt{xyz}=xy+yz+zx \implies xyz=(xy+yz+zx)^2 \geq 3xyz(x+y+z) \implies x+y+z\leq \frac{1}{3}$