# Showing $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$ for $x,y,z \ge 0$

How do I show that $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$ for $x,y,z \ge 0$?

Let $A = x+y+z$, then $z=A-x-y$. Maximize $f(x,y)=xy(A-x-y)$ on region enclosed by $x=y=0$ and $x+y=A$

What I don't understand is the idea behind this method.

• Why $A=x+y+z$ not $A=\frac{x+y+z}{3}$?
• I guess $f(x,y)=xy(A-x-y)$ is to see whats the maximum of the equation
• But why regions $x=y=0$ and $x+y=A$?

Sorry this seems like an easy question but somehow I don't get it ...

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For your first question, the real answer is that it doesn't matter. We could just as well let $A=\frac{x+y+z}{3}$. Then $z=3A-x-y$, and so on. The calculations end up very similar. –  André Nicolas Apr 7 '12 at 14:45
An equivalent problem is to show (e.g. by Lagrange's method ) that the maximum value of $x^{2}y^{2}z^{2}$ subject to $x^{2}+y^{2}+z^{2}=R^{2}$ is $\left( R^{2}/3\right) ^{3}$, from which follows the inequality $(a_{1}a_{2}a_{3})^{1/3}\leq (a_{1}+a_{2}+a_{3})/3$, where $a_{1},a_{2},a_{3}$ are positive numbers. –  Américo Tavares Apr 7 '12 at 20:53

Don't feel bad that you think things are confusing; In my opinion, this is a rather subtle problem for a second year calculus course.

It is the right hand side of the inequality being considered here. The maximum value of $xyz$ is calculated in the special case that the sum $x+y+z$ is known: the formulation given in the hint is finding the maximum value of the function $$\tag{1}g(x,y,z)=xyz$$ subject to to constraints $x\ge0,y\ge0,z\ge0$ and $x+y+z=A$, where $A$ is a fixed, but otherwise arbitrary, number. The last constraint forces $z=A-x-y$, which allows us to phrase the maximization problem as

Find the maximum value of $$\tag{2} f(x,y)=xy(A-x-y)$$ over the region defined by $x,y\ge0$, $A-x-y\ge 0$. Note that that last inequality can be written as $A\ge x+y$. This region can be described as given in the hint "the region enclosed by $x=0$, $y=0$, and $x+y=A$".

Why do this? Well, .... umm ... it just works:

So, you go about finding the gradient of $f$, setting it equal to the zero vector, finding the maximum value of $f$ over those zeroes and at the boundary of the region, and then find the largest of those maximums and where it occurs. (I can provide details here if you like, but the calculations aren't too bad.)

You will find that the maximum value of $f$ over the given region is attained at $x=y={A\over3}$. Thus, the maximum value of $g$ is attained at $x=y=z={A\over 3}$ (remember $A=x+y+z)$ and this maximum value is $$\textstyle g\bigl({A\over3},{A\over3},{A\over3}\bigr)={A\over 3}\cdot{A\over3}\cdot{A\over3}={A^3\over 27}.$$

So, what have we shown?

We have shown the following:

If $x,y,z\ge0$ and if $x+y+z=A$ (where $A$ is now just the sum of the given $x,y,z$), then $xyz\le {A^3\over 27}$. Thus, we have $$xyz\le{A^3\over27}={(x+y+z)^3\over 27};$$ and taking cube roots of both sides of the above gives your inequality (which, by the way, is a very important and famous inequality known as the Arithmetic Mean-Geometric Mean, or AM-GM, inequality).

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For your first question, the answer is that it doesn't matter. We could just as well let $A=\frac{x+y+z}{3}$. Then $z=3A-x-y$, and so on. The calculations end up very similar. Indeed things are a little prettier, I think, if we let $\frac{x+y+z}{3}=A$. But it makes no real difference.

The point is that we fix $x+y+z$, and show using tools from the calculus that for the fixed value $x+y+z=A$, we must have $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$. Since our argument works for any non-negative $A$, that shows that the desired inequality always holds.

In order to show that $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$, we find the maximum value of $xyz$, given that $x+y+z=A$ and $x,y,z\ge 0$. Since $z=A-x-y$, we want to find the maximum value of $f(x,y)=xy(A-x-y)$. It will turn out when you maximize that this maximum value is $A^3/27$, which gives exactly the right result. For it implies that the maximum of $\sqrt[3]{xyz}$ given that $x+y+z=A$ is $\sqrt[3]{A^3/27}$, which is $A/3$.

Of course we will only consider $x,y\ge 0$. The condition $z\ge 0$ becomes, in terms of $x$ and $y$, that $A-x-y \ge 0$. This can be rewritten as $x+y \le A$.

So we are maximizing the function (not equation) $xy(A-x-y)$ in the region that satisfies $x\ge 0$, $y\ge 0$, $x+y \le A$.

This is the same problem as maximizing $xy(A-x-y)$ over the triangle bounded by the lines $x=0$, $y=0$, and $x+y=A$. (The condition $x+y \le A$ just says that $(x,y)$ lies below the line $x+y=A$.)

I am sure you can handle the partial derivatives stuff.

Remark: Actually, because the inequality we are interested in is homogeneous (it holds at $(x,y,z)$ iff it holds at $(tx,ty,tz)$), we could even assume that $x+y+z$ is some specific number, like $3$.

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This is AM-GM inequality in $3$ variables. Read the complete solution of Exercise 1 from Dr. Titus notes

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For the first question, $A$= $x+$y$+z$ since it is being used as a substitution in the equation; if it was written with being divided by 3, it wouldn't be an "appropriate" substitution in the sense that the whole equation would be wrong. Yes, $f(x,y)$ must equal $xy(A-x-y)$ in order for the maximum to be found(Lagrange multipliers), and for the final question, it is for the same reason as the first.

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