hello, we get by Lagrange Multipliers $$x+y+z\le \sqrt{3}$$ and the equal sign holds if $$x=z=1,y=-2+\sqrt{3}$$ Sonnhard.

Equality holds when $\{x,y,z\}=\{-1,-1,2+\sqrt{3}\}.$

Actually, you can't use $\{x,y,z\}=\{1,1,-2+\sqrt{3}\}$, because $-2+\sqrt{3}\in(-1,1)$.

Yeah. Thank Radar.

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ok , i have another proof here is what , clearly $x,y,z$ can't have the same sign if only $x$ is negative , let $f(x)=x+\frac{1}{x}$ so $f''(x)=\frac{2}{x^3}$ since $y,z>0$ we get (let$ x=-x'$) $f(x')=f(y)+f(z) \ge 2f(\frac{y+z}{2})$ clearly the max of $x+y+z$ is positive , so we have $y+z >x'$ and since $x'>2, \frac{y+z}{2}>1$ and we have $f(a)>f(b)$ if $a>b>1$ so $f(x')> 2f(\frac{x'}{2})$ so $\frac{x'^2+1}{x'} > \frac{x'^2+4}{x'}$ contradiction , so we have $f(x)=f(y)+f(z)$ and we want $max(x-y-z)$ we have $x^2+1-x(f(y)+f(z))=0$ take $x$ as the biggest root of above , so $x=\frac{f(y)+f(z)+\sqrt{(f(y)+f(z))^2-4}}{2} $ so we want $max \frac{1}{z}-z+\frac{1}{y}-y +\sqrt{((f(y)+f(z))^2-4)}$(we just took away the $2$ ) let $y \ge z \ge 1$ we use derivative with respect to $y$ to show this functions is infact decreasing if we take $y$ higher , so we prove the derivative is always negative the derivative is (note : we write every thing in terms of $f(y),f(z)$ $-\frac{f(y)}{y}+\frac{(2-\frac{f(y)}{y})}{\sqrt{1- \frac{4}{(f(y)+f(z))^2}}}$ since $f(z)\le f(y)$ we just assume $1=z$ for things to go better we must say $\frac{f(y)}{y} \ge \frac{(2-\frac{f(y)}{y})(f(y)+2)}{\sqrt{f(y)^2+4f(y)}}$ clearing some things we should prove $\frac{f(y)^3+4f(y)^2}{f(y)+2}>2y-f(y)$ which is clear since $\frac{f(y)^3+f(y)^2}{f(y)+2} \ge f(y) > 2y-f(y)$ so we can let $y=z$ in the original problem state ment so we want $$max (\frac{1}{y}-y+\sqrt{y^2+1+\frac{1}{y^2}})$$taking derivative of $y$ we show its $max$ is in $y=1$ we just prove that the above function is decreasing in terms of $y$ the derivative is $\frac{y-\frac{1}{y^3}}{\sqrt{y^2+1+\frac{1}{y^2}}} -1-\frac{1}{y^2}$ but since $\frac{y}{y^2+1+\frac{1}{y^2}} \le 1$ we are done so we can just take $y=1$ so we get $$max(x+y+z)=\sqrt{3}$$for $x=2+\sqrt{3},y=z=-1$ and other permutations

Radar wrote: Real numbers $x,y,z$ satisfy $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+x+y+z=0$$and none of them lies in the open interval $(-1,1)$. Find the maximum value of $x+y+z$. Proposed by Jaromír Šimša This is a high class problem.

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