Exercise 1.2.32

Wikipedia gives this proof of geometric mean $G=\sqrt[3]{\mathit{xyz}}\le$ arithmetic mean $A=(x+y+z)∕3$. First there is equality in case $x=y=z$. Otherwise $A$ is somewhere between the three positive numbers, say for example $z<A<y$.

Use the known inequality $g\le a$ for the two positive numbers $x$ and $y+z-A$. Their mean $a=\frac{1}{2}(x+y+z-A)$ is $\frac{1}{2}(3A-A)=$ same as $A!$ So $a\ge g$ says that $A^3\ge g^{2}A=x(y+z-A)A$. But $(y+z-A)A=(y-A)(A-z)+\mathit{yz}>\mathit{yz}$ Substitute to find $A^3>\mathit{xyz}=G^3$ as we wanted to prove. Not easy!

There are many proofs of $G={\left(x_1{x}_2\cdots x_n\right)}^{1∕n}\le A=\left(x_1+x_2+\cdots +x_n\right)∕n$. In calculus you are maximizing $G$ on the plane $x_1+x_2+\cdots +x_n=n$. The maximum occurs when all $x$ ’s are equal.