Olympiad problem

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Let the expression above be $P$ .

By Cauchy-Schwarz inequality,

$P=(x+y+z)(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq (x+y+z){(\sqrt{1}+\sqrt{1}+\sqrt{1})}^{2}=9(x+y+z) \geq \frac{27}{2}$ .

We check that this minimum is attained when $x=y=z=\frac{1}{2}$ .

Next, viewing the expression as a function in $x$ (by writing $C=y+z$ and $D=\frac{1}{y}+\frac{1}{z}$ respectively) gives:

$P={(x+c)}^{2}(\frac{1}{x}+D)$

$=Dx^{2}+(2CD+1)x+\frac{C^{2}}{x}+C^{2}D+2C$

Note that

$P'(x)=2Dx+2CD+1-\frac{C^{2}}{x^{2}}$

$P''(x)=2D+\frac{2C^{2}}{x^{3}}$

Since all $x,C$ and $D$ are positive, $P''(x)>0$ , therefore $P(x)$ is a convex function.

Viewing the expression as functions of $y$ and $z$ , we arrive at the similar conclusion that the expression is convex in terms of $x,y$ and $z$ . Since the maximum of a positive convex function is attained when all its variables are at the extremum of their respective domains. Checking the value of $P$ when $x,y,z \in \{\frac{1}{2},1\}$ , we find that maximum is attained when $x=y=z=1$ .

Upon some computation, the minimum is $\frac{27}{2}$ while the maximum is $27$ . Adding them up yields $\boxed{40.5}$ , hence the answer.