How can I prove that if $$ (x+y+z-3)^2=xy+yz+zx-3, \ x,y,z\in\mathbb{R} $$ Then $$ 3+ xyz(x+y+z)\geq 6xyz $$ without using Lagrange multipliers?
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Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative and $xyz=w^3$.
Hence, the condition does not depend on $w^3$ and we need to prove that $f(w^3)\geq0$, where $f$ is a linear function.
But a linear function gets a minimal value for an extremal value of $w^3$,
which happens for equality case of two variables.
Id est, it remains to prove our inequality for $y=x$ and $(2x+z-3)^2=x^2+2xz-3$.
We need to prove that $3+x^2z(2x+z)\geq6x^2z$ and the rest is easy.
Michael Rozenberg
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In this case $v^2\ge 1$ from the condition, so that isn't a complication. However proving extrema of $w^3$ occurs when two among the variables are equal may need Lagrange or a more complicated theorem than the question itself. – Macavity Mar 07 '16 at 07:03
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@ Macavity $x$, $y$ and $z$ are real roots of the equation $X^3-3uX^2+3v^2X=w^3$. Draw graph of $f(X)=X^3-3uX^2+3v^2X$ and a line $Y=w^3$ and you'll see that it's obvious (about occurring of extremal value of $w^3$). – Michael Rozenberg Mar 07 '16 at 10:45
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I am aware of the $u, v, w$ method and its proof (without resorting to visualizing multidimensional graphs) - I don't think its as trivial as you make it though. There must be simpler solutions. – Macavity Mar 08 '16 at 06:16
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@Macavity draw a cube parabola and a parallel line to OX and I am sure that you'll see that it's trivial. It's indeed trivial. – Michael Rozenberg Mar 08 '16 at 13:45