Inequality problems

enndb0x

The following problem was proposed in Inequalities Marathon at Pre Olympiad forum

Problem 138: Let $x,y,z > 0$ such that

Prove that
$(x - 1)(y - 1)(z - 1) \le 6 \sqrt {3} - 10$

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This was my solution ,and it is wrong but I dunno why ,hope you will find the mistake.

solution

**Posted:** Mon Nov 02, 2009 4:41 pm

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Inequalities Marathon

JBL

**Posted:** Mon Nov 02, 2009 5:19 pm

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Joel
Hi Deeps! <3

Kent Merryfield

Since you posted it in the calculus section, we might as well look at the Lagrange multipliers solution. (General hint: when there are many circumstances, notably including symmetry across multiple variables, in which Lagrange multipliers is far nicer than one-variable calculus techniques.)

Maximize $f(x)=(x-1)(y-1)(z-1)$ subject to $g(x,y,z)=xyz-x-y-z=0.$

We set $\nabla f=\lambda\nabla g.$

$(y-1)(z-1)=\lambda(yz-1)$
$(x-1)(z-1)=\lambda(xz-1)$
$(x-1)(y-1)=\lambda(xy-1)$

Solve the first and second equations together:

$\frac{(y-1)(z-1)}{yz-1}=\frac{(x-1)(z-1)}{xz-1}$

Either $z=1$ or $(y-1)(xz-1)=(x-1)(yz-1)$

$xyz-xz-y+1-xyz-yz-x+1$

$xz-x=yz-y,$ so dividing by $z-1$ again, we have $x=y.$

Similar work gives us that either $x=1$ or $y=z$ .

If none of $x,y,$ or $z$ are $1,$ then we have $x=y=z.$ Put that into $g(x,y,z)=0$ to get that $x^3-3x=0,$ so $x=0$ or $x=\pm\sqrt{3}.$

We have the following points that are either critical point or points that we haven't ruled out yet:

$x=y=z=-\sqrt{3}$ in which case $f(x,y,z)<0.$

$x=y=z=0,$ in which case $f(x,y,z)=-1.$

An unknown number of points for which at least one of $x,y,$ or $z$ equals $1,$ in which case $f(x,y,z)=0.$

$x=y=z=\sqrt{3},$ in which case $f(x,y,z)=6\sqrt{3}-10.$

We're still not done. The constraint set isn't compact (isn't bounded), so we still have to do something else to assure that we've found the global maximum of $f.$

**Posted:** Mon Nov 02, 2009 10:21 pm