prove that $xy+yz+zx\ge x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}$

Question

prove that $xy+yz+zx\ge x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}$ if $x,y,z>0$

My try : dividing inequality by $\sqrt{xyz}$ and putting $\sqrt{x}=a,\sqrt{y}=b,\sqrt{z}=c$

we have to prove $$\sum_{cyc}\frac{ab}{c}\ge a+b+c$$ or $$2\sum_{cyc}\frac{ab}{c}\ge 2(a+b+c)$$ using $\frac{ab}{c}+\frac{bc}{a}\ge 2b$ and similarly for others the proof can be completed.

Is it correct? Also i am looking for different proofs for this (possibly more simpler ).Thanks

Answer 1

Yes, it's correct.

Also, SOS helps: $$\sum_{cyc}(xy-x\sqrt{yz})=\frac{1}{2}\sum_{cyc}(xz+yz-2z\sqrt{xy})=\frac{1}{2}\sum_{cyc}z(\sqrt{x}-\sqrt{y})^2\geq0.$$

Answer 2

$(xy+yz+zx)(zx+xy+yz)\geq( x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy})^2$

by Cauchy Inequality.

Answer 3

$x,y,z>0$ and that the inequality is a symmetric expression implies that we can take without loss of generality, an ordering $x\ge y\ge z \implies xy\ge zx\ge yz \implies \sqrt{xy}\ge \sqrt{zx}\ge \sqrt{yz}$

so that the sequences $\{\sqrt{xy},\sqrt{zx},\sqrt{yz}\}, \{\sqrt{xy},\sqrt{zx},\sqrt{yz}\}$, (i.e. the same sequence) are similarly sorted, so that, by the Rearrangement Inequality, we have

$$\sqrt{xy}\sqrt{xy}+\sqrt{yz}\sqrt{yz}+\sqrt{zx}\sqrt{zx}\ge \sqrt{xy}\sqrt{zx}+\sqrt{yz}\sqrt{xy}+\sqrt{zx}\sqrt{yz}$$ which is the required inequality.

Answer 4

$$xy+yz+zx = {xy+yz\over 2} + {yz+zx\over 2} + {zx+xy\over 2} \geq x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}$$ By AM-GM Inequality. Done!