1
$\begingroup$

How to find equality of maximum or minimum value of the given expression? I understood that if expression and condition are symmetric, then we may assume that all of the variables are equal. But how to know at least one of the expression and condition is not symmetric?

For example, $x,y,z\in\mathbb{R^{+}}$ and $3x^2+4y^2+5z^2=2xyz$ are given. Then how to find equality the minimum value of this expression $3x+2y+z$?

Improve this question
$\endgroup$
  • 2
    $\begingroup$ Check out Lagrange Multipliers, which is used to find local extrema of a function given a constraint equation. $\endgroup$ – DEATH_CUBE_K May 9 '20 at 5:42
  • 1
    $\begingroup$ In fact if the expression is symmetrical, it doesn't mean minimum or maximum has to be reached when all variables are equal. That understand is not always correct, though often it seems that way for simple functions. $\endgroup$ – Macavity May 9 '20 at 7:00
  • $\begingroup$ @DEATH_CUBE_K Ok thatbk you $\endgroup$ – Mutse May 9 '20 at 22:19
  • $\begingroup$ @DEATH_CUBE_K But it doesn't work for all problems. For example USAMO 2017/6. You can see problem here: artofproblemsolving.com/community/c5h1434574p8117097 $\endgroup$ – Mutse May 9 '20 at 22:21
  • $\begingroup$ @Macavity Can you give me an example? Do you know when it's not correct? $\endgroup$ – Mutse May 9 '20 at 23:58
2
$\begingroup$

By AM-GM $$3x+2y+z=\frac{(3x^2+4y^2+5z^2)(3x+2y+z)}{2xyz}\geq$$ $$\geq\frac{12\sqrt[12]{(x^2)^3(y^2)^4(z^2)^5}\cdot6\sqrt[6]{x^3y^2z}}{2xyz}=36.$$ The equality occurs for $x=y=z$ and $3x^2+4y^2+5z^2=2xyz,$ which says that we got a minimal value.

Also, you can use the following way.

Let $f(x,y,z,\lambda)=3x+2y+z+\lambda(3x^2+4y^2+5z^2-2xyz).$

Thus, in the minimum point we have $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=\frac{\partial f}{\partial \lambda}=0$$ and you'll get the system, which gives that $(6,6,6)$ is a critical point and by using second partial derivatives we can get that it's a minimum point.

I think, it's better to look for the first way before.

Improve this answer
$\endgroup$
  • $\begingroup$ I am wondering how were you able to come up with such sophisticated way of using AM-GM? $\endgroup$ – Shiv Tavker May 9 '20 at 22:03
  • $\begingroup$ Sorry for that just saw that you have answered almost every question that has AM GM tag. Is this a general technique to multiply the objective function with constraints to check if AM-GM will be useful here? $\endgroup$ – Shiv Tavker May 9 '20 at 22:07
  • $\begingroup$ Thank you for a nice solution! But your both ways don't work on problems with fractional expressions(for example, $\sum_{cyc}\frac{a}{b^3+4}$). So do you have more idea(method, advice) on this type problem? $\endgroup$ – Mutse May 9 '20 at 23:45
  • $\begingroup$ @Shiv Tavker In this type inequalities we always can find solution by AM-GM, The problem is the case of the equality occurring. It can be a very ugly point and and the solution would be very ugly. Because we need to find this point before. $\endgroup$ – Michael Rozenberg May 10 '20 at 3:29
  • 1
    $\begingroup$ @Mutse The general method for proving inequalities does not exist. But If you post some inequality I can try to prove it. $\endgroup$ – Michael Rozenberg May 10 '20 at 3:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.