# Maximum of given expression?

Suppose $a,b,c>0$ and further that $a^{2} + b^{2} + c^{2}=2abc + 1$.

The problem is to find $\max \big(a-2bc\big) \big(b-2ca\big) \big(c-2ab\big)$.

Give me some help. I've tried $X=a-2bc$, $Y=b-2ca$, $Z=c-2ab$ which yields $X^2 + Y^2 + Z^2 = 1-2XYZ$, but $\frac12$ is not the maximum because $XYZ=0$.

Can someone give me an elegant solution?

• Please see this tutorial on how to use mathjax to make your mathematics legible. – Mike Pierce May 29 '15 at 14:14

let $$x=a-2bc,y=b-2ca,z=c-2ab$$ since $$a^2+b^2+c^2=2abc+1\Longrightarrow x^2+y^2+z^2+2xyz=1$$ so we only find $xyz$ maximu,since $$1=x^2+y^2+z^2+2xyz\ge 3(x^2y^2z^2)^{\frac{1}{3}}+2xyz\Longrightarrow xyz\le\dfrac{1}{8}$$ so $$(a-2bc)(b-2ac)(c-2ab)\le\dfrac{1}{8}$$
HINT: with the method of Lagrange Multipliers we get $$(a-2bc)(b-2ca)(c-2ab)\le \frac{1}{8}$$ and the equal sign holds if $$a=b=\frac{1}{2},c=1$$
This is what the method of Lagrange Multipliers was made for. You have a function of three variables $$f(a,b,c) = (a-2bc)(b-2ca)(c-2ab)$$ and a constraint $$g(a,b,c)=a^2 + b^2 + c^2 -2abc - 1 = 0.$$
The constraint tells us that we are looking for the maximum on a surface defined by that equation. If such a maximum exists, then the level curves of $f(a,b,c)$ must be parallel to the surface given by $g(a,b,c)=0$. This means that the gradients of the curves at that point are also parallel. That is, there exists a nonzero $\lambda$ such that $$\nabla f(a,b,c) = \lambda \nabla g(a,b,c).$$