Inequality about symmetric mean ($8(x+y+z)(xy+yz+xz)\leq 9(x+y)(y+z)(x+z)$)

Question

I have three question about this :

(1) I hope to show the inequality above, for another competition problem. I have tried to make these symmetric terms into a cubic equation, and try its discriminant but failed.

(2)I also found some Olympiad resourse. It is only applied to show another inequality and is said well-known. However, does it have a name ?

(3)For a general case, since we can rewrite it : $\frac{xy+yz+xz}{3} \frac{x+y+z}{3} \leq \frac{x+y}{2} \frac{y+z}{2} \frac{x+z}{2}$. Is there any research about inequalities about product of different means of a given $k$ number??

Answer 1

Assuming $x,y,z$ are non negative reals ... \begin{eqnarray*} x(y-z)^2+y(z-x)^2+z(x-y)^2 \geq 0 \end{eqnarray*} and rearrange.

Answer 2

By expansion, it can verified that the remarkable identity $$(x+y+z)(xy+yz+zx) = (x+y)(y+z)(z+x)+xyz$$ holds. So your inequality is equivalent to $$8((x+y)(y+z)(z+x)+xyz)\le 9(x+y)(y+z)(z+x),$$ which simplifies to $$8xyz\le (x+y)(y+z)(z+x).$$ By AM-GM or simply the trivial inequality, $$\frac{x+y}{2}\ge \sqrt{xy}.$$ Multiplying the three instances of this yields the desired inequality.