<s>[INST] Show $2(x+y+z)-xyz\leq 10$ if $x^2+y^2+z^2=9$ If $x,y,z$ are real and $x^2+y^2+z^2=9$, how can we prove that $2(x+y+z)-xyz\leq 10$?
Please provide a solution without the use of calculus. I know the solution in that way.
 [/INST] Here is an attempt. I believe it has to be reconsidered for negative numbers, though.
$0\leq (x+y+z-1)^2$
$0\leq x^2+y^2+z^2+1-2(x+y+z)+2xyz(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$
$2(x+y+z)-2xyz(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\leq x^2+y^2+z^2+1$
EDIT: As pointed out my bound is the wrong way :( But I hope the first steps gives others some ideas.
That's almost what we want. We only need to bound the harmonic mean by the quadratic mean now.
$\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\leq \sqrt{\frac{x^2+y^2+z^2}{3}}$
In our case:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \sqrt{3}$
which is enough.
Hope there is no mistake :)
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