If the sum and the product of three positive numbers $x,y,z$ be respectively 8 and 18 then prove that $p\in[144,\frac{8^4}{27}]$ where $p=(8-x)(8-y)(8-z)$.
I got the max value of $p$ by AM-GM inequality but could not get minimum value.
Please help me.
Since $x+y+z=8$, we have: $$ p=(x+y)(y+z)(z+x)\ge(2\sqrt{xy})(2\sqrt{yz})(2\sqrt{zx})=8xyz=144 $$ Since $xyz=18$.
But note, that the lower and upper bound cannot be achieved, because both require $x=y=z$ which implies $x=2$ and $x=18^{\frac{1}{3}}$ at the same time.