Prove that for a,b,c > 0, geometric mean <= arithmetic mean

[EnlightenedOne]

Homework Statement

Let $a,b,c \in \mathbb{R}^{+}$.

Prove that $$ \sqrt[3]{abc} \leq \frac{a+b+c}{3}. $$
Note: $a,b,c$ can be expressed as $a = r^3, b = s^3, c = t^3$ for $r,s,t > 0$.

Homework Equations

$P(a,b,c): a,b,c \in \mathbb{R}^{+}$

$Q(a,b,c): \sqrt[3]{abc} \leq \frac{a+b+c}{3}$

The Attempt at a Solution

On the side, I tried to substitute $a = r^3, b = s^3, c = t^3$ to see where it would take me, and I ended up here:

$r^3 + s^3 + t^3 -3rst \geq 0$

So, in order to directly prove the original statement, I would first have to start with a statement involving $r,s,t \geq 0$ and then work backwards to $Q(a,b,c)$.

In other words, before I begin my proof I need to show that $r^3 + s^3 + t^3 -3rst \geq 0$, but in order to do that I would have to factor it somehow.

The solution says:
"Observe that $r^3 + s^3 + t^3 -3rst = \frac{1}{2} (r+s+t)[(r-s)^2 + (s-t)^2 +(t-r)^2]$"

So, my question is:
How would I have factored $$ r^3 + s^3 + t^3 -3rst $$ to $$ \frac{1}{2} (r+s+t)[(r-s)^2 + (s-t)^2 +(t-r)^2] $$ without using software? (paper and pencil) It seems impossible that I would have ever been able to factor that.

Thanks!

 

[RUber]

It is tough to speculate about how one might know how to factor a polynomial. Now that you have seen it, you will likely think to do it next time you see a similar structure.

However, you could attack the proof in a variety of ways, since if it true, it doesn't matter how you show it, since it will always be true.

What if you looked at the problem like:

Without loss of generality, assume $a\leq b \leq c$
abc= $x(x-y)(x+z)= x^3 + (z-y)x^2 - xyz$
where x=b, x-y = a, and x+z = c.
Since all the values are non-negative, you also know that $y\leq x$.
And:
$\left( \frac{ a+b+c }{3} \right)^3 = \frac{a^3 + b^3 + c^3 +3a^2b+3a^2c + 3b^2a+3b^2c + 3 c^2 a+3c^2 b + 6abc}{27}$
can be replaced with:
$\left( \frac{ 3x-y+z }{3} \right)^3 = \frac{27x^3 - y^3 + z^3 -27x^2y+27x^2z + 9y^2x+3y^2z + 9 z^2 x-3z^2 y - 18xyz}{27}$
Then we need to show:
$x^3 + (z-y)x^2 - xyz \leq \frac{27x^3 - y^3 + z^3 -27x^2y+27x^2z + 9y^2x+3y^2z + 9 z^2 x-3z^2 y - 18xyz}{27}$

 

[Ray Vickson]

Homework Statement

Let $a,b,c \in \mathbb{R}^{+}$.

Prove that $$ \sqrt[3]{abc} \leq \frac{a+b+c}{3}. $$
Note: $a,b,c$ can be expressed as $a = r^3, b = s^3, c = t^3$ for $r,s,t > 0$.

Homework Equations

$P(a,b,c): a,b,c \in \mathbb{R}^{+}$

$Q(a,b,c): \sqrt[3]{abc} \leq \frac{a+b+c}{3}$

The Attempt at a Solution

On the side, I tried to substitute $a = r^3, b = s^3, c = t^3$ to see where it would take me, and I ended up here:

$r^3 + s^3 + t^3 -3rst \geq 0$

So, in order to directly prove the original statement, I would first have to start with a statement involving $r,s,t \geq 0$ and then work backwards to $Q(a,b,c)$.

In other words, before I begin my proof I need to show that $r^3 + s^3 + t^3 -3rst \geq 0$, but in order to do that I would have to factor it somehow.

The solution says:
"Observe that $r^3 + s^3 + t^3 -3rst = \frac{1}{2} (r+s+t)[(r-s)^2 + (s-t)^2 +(t-r)^2]$"

So, my question is:
How would I have factored $$ r^3 + s^3 + t^3 -3rst $$ to $$ \frac{1}{2} (r+s+t)[(r-s)^2 + (s-t)^2 +(t-r)^2] $$ without using software? (paper and pencil) It seems impossible that I would have ever been able to factor that.

Thanks!

Hints: (1) the function $\ln(x)$ is concave in $x > 0$; (2) Jensen's inequality.

In fact, it is just as easy to prove that for $x_1, x_2, \ldots, x_n, w_1, w_2, \ldots, w_n \in \mathbb{R}^{+}$ with $\sum_{i=1}^n w_i = 1$ we have
$$w_1 x_2 + w_2 x_2 + \cdots + w_n x_n \geq x_1^{w_1} x_2^{w_2} \cdots x_n^{w_n},$$
with equality iff $x_1 = x_2 = \cdots = x_n$.

 

[wabbit]

Honestly I understand your perplexity. There are far more natural ways to prove this, in a way that is straightforward to generalize to any number of variables, than to use this rather special identity (I'm far too lazy to even check if that identity is actually true). Suggestions as to how to do this have been provided by others already.

That is, unless you're doing a course on symmetric polynomials - then the method might seem quite natural).

 

[EnlightenedOne]

Thanks for all the replies guys, I appreciate it! I am actually in a Foundations of Mathematics course (Intro to proofs) and preparing for Advanced Calculus that I'm taking next Fall, so most of the suggestions you guys have made are probably way above my level :D, but thanks anyway! Really, this was just in a homework set under "Proofs for Real Numbers", and this is my first proof course. I already know how my textbook wants me to do the proof, but I just had trouble figuring out how they factored the expression. They are factoring it because once it can be shown that r^3 + s^3 + t^3 -3rst is always >= 0, then you can start with this expression and work backwards to Q's expression, thus giving a direct proof of the original result.

 

[wabbit]

Ah I see... Yes you should be able to prove that identity by hand : just take a large sheet of paper, brace yourself, expand the terms and express the RHS as a sum of terms of the form $r^k t^l s^m$ - the identity will show up if true.

What I find difficult is what you mentionned first, how to guess that identity in the first place? To me this sounds quite demanding for a first course in proofs :)

 

[vela]

The solution says:
"Observe that $r^3 + s^3 + t^3 -3rst = \frac{1}{2} (r+s+t)[(r-s)^2 + (s-t)^2 +(t-r)^2]$"

So, my question is: How would I have factored $r^3 + s^3 + t^3 -3rst$ to $\frac{1}{2} (r+s+t)[(r-s)^2 + (s-t)^2 +(t-r)^2]$ without using software (paper and pencil)? It seems impossible that I would have ever been able to factor that.

You futz around with the expression, trying different things, until it works out or until you give up. In this case, those terms suggest looking at the expansion of $(r+s+t)^3$. This leads to
$$r^3+s^3+t^3-3rst = (r+s+t)^3 - 3(r^2 s + r s^2+r^2 t+r t^2+s^2 t+s t^2+3 r s t).$$ Then you try factor that last part, using symmetry as a guide. You might hope that (r+s+t) is a factor, so you could try factoring out "extra" factor of r, s, and t from various terms. Eventually, you'd find
$$r^2 s + r s^2+r^2 t+r t^2+s^2 t+s t^2+3 r s t = rs(r+s+t) + rt(r+t+s) + st(s+t+r) = (rs+rt+st)(r+s+t).$$ So now you have
$$r^3 + s^3 + t^3 -3rst = (r+s+t)^3 - 3(r+s+t)(rs+rt+st) = (r+s+t)[(r+s+t)^2 - 3(rs+rt+st)].$$ And so on. It's really just intuition, educated guesses, and hoping for the best.

Of course, when you write your proof out, you might decide to not show all this scratch work. Instead you start off with "Observe that..." and pull the result seemingly out of thin air.

 

[EnlightenedOne]

@wabbit I agree, and feel the same way :D. I actually did prove the identity to myself, and it was very tedious... I just hope that proofs like these don't show up on a test, because I'm just going to skip it if it is not this exact same proof (because now I know the answer xD). There was actually a question I did earlier that was very similar to this one, except that it involved just 2 positive real numbers a,b instead of 3, and I was actually able to do that one (because all of the factoring was quadratic instead of cubic).

@vela Ok, that's what I thought... You play around with it and hope that you get somewhere, which I hate doing because it (without looking at the solution) seems like I'm just wasting time and getting nowhere. I still don't think I would have actually arrived at that factorization within any reasonable time frame (perhaps never). I see what you did though.

Thanks again everyone!

 

[Ray Vickson]

Thanks for all the replies guys, I appreciate it! I am actually in a Foundations of Mathematics course (Intro to proofs) and preparing for Advanced Calculus that I'm taking next Fall, so most of the suggestions you guys have made are probably way above my level :D, but thanks anyway! Really, this was just in a homework set under "Proofs for Real Numbers", and this is my first proof course. I already know how my textbook wants me to do the proof, but I just had trouble figuring out how they factored the expression. They are factoring it because once it can be shown that r^3 + s^3 + t^3 -3rst is always >= 0, then you can start with this expression and work backwards to Q's expression, thus giving a direct proof of the original result.

@wabbit I agree, and feel the same way :D. I actually did prove the identity to myself, and it was very tedious... I just hope that proofs like these don't show up on a test, because I'm just going to skip it if it is not this exact same proof (because now I know the answer xD). There was actually a question I did earlier that was very similar to this one, except that it involved just 2 positive real numbers a,b instead of 3, and I was actually able to do that one (because all of the factoring was quadratic instead of cubic).

@vela Ok, that's what I thought... You play around with it and hope that you get somewhere, which I hate doing because it (without looking at the solution) seems like I'm just wasting time and getting nowhere. I still don't think I would have actually arrived at that factorization within any reasonable time frame (perhaps never). I see what you did though.

Thanks again everyone!

Are you absolutely required to do the proof the way the book described? There are easier ways that do not involve tools you do not yet possess. Here is one.

Note that both sides of your alleged inequality scale in the same way, so that if it is true (or false) for $a,b,c$ it is equally true (or false) for $a'=ka, b' = kb, c' = kc$ for any constant $k > 0$. Therefore, we might as well assume that $a+b+c = 1$, because we can always get that by re-scaling the problem. So, now you want to prove that
$$a,b,c > 0,\: a+b+c=1\; \longrightarrow 1/3 \geq \sqrt[3]{abc}, \: \text{or} \; 1/27 \geq abc$$

We need to know the maximum of $f(a,b,c) = abc$ on $S_3 = \{ (a,b,c)| a,b,c > 0, a+b+c=1\}$. On $S_3$ we have $c = 1 - a - b$, so we need to find the maximum of $F(a,b) = f(a,b,1-a-b) = ab(1-a-b)$ on $S_2 = \{(a,b) | a>0,b>0,a+b<1 \}$. The function $F \to 0$ as $(a,b) \to$ the boundary of $S_2$, and $F > 0$ inside $S_2$, so the maximum will occur at a stationary point inside $S_2$. We can just set the two partial derivatives of $F$ to zero and solve for $a,b$. That gives $a = b = 1/3$, so $c = 1/3$, hence $f \leq 1/27$ on $S_3$, with equality at $a = b = c = 1/3$.

 

[EnlightenedOne]

@Ray Vickson No, we are not required to do it the way the book does it, but there is a certain structure to the types of proofs in this section of the book and we are implicitly supposed to follow that structure because we are not experienced with proofs yet. That's an interesting approach, some of which I don't quite understand. I understand most of what you are doing (the calculus parts), but things like setting a+b+c=1 etc., I'm not sure I fully understand. Thanks for the method though, as I have now see many ways to prove this xD. But, really, the only thing I needed to show was that r^3 + s^3 + t^3 - 3rst >= 0 (because I actually got to this expression from the original Q statement), so that I could start with this expression ("pulling it out of the air", so to speak) from the assumption that a,b,c > 0 , and then expand it to reach the original expression, i.e. geometric mean <= arithmetic mean.

 

[Ray Vickson]

@Ray Vickson No, we are not required to do it the way the book does it, but there is a certain structure to the types of proofs in this section of the book and we are implicitly supposed to follow that structure because we are not experienced with proofs yet. That's an interesting approach, some of which I don't quite understand. I understand most of what you are doing (the calculus parts), but things like setting a+b+c=1 etc., I'm not sure I fully understand. Thanks for the method though, as I have now see many ways to prove this xD. But, really, the only thing I needed to show was that r^3 + s^3 + t^3 - 3rst >= 0 (because I actually got to this expression from the original Q statement), so that I could start with this expression ("pulling it out of the air", so to speak) from the assumption that a,b,c > 0 , and then expand it to reach the original expression, i.e. geometric mean <= arithmetic mean.

If $$(a'+b'+c')/3\; [\geq , \leq ]\; \sqrt[3]{a' b' c'},$$ then for any $k > 0$ we have $$k(a'+b'+c')/3 = (ka' + kb' + kc')/3\; [\geq , \leq ]\; k \sqrt[3]{a' b' c'} = \sqrt[3]{ka' \cdot kb' \cdot kc'}$$ We can always take $k$ to give us $(ka' + kb' + kc') = 1$, then call $ka', kb', kc'$ by the new names $a,b,c$.

 

[EnlightenedOne]

If $$(a'+b'+c')/3\; [\geq , \leq ]\; \sqrt[3]{a' b' c'},$$ then for any $k > 0$ we have $$k(a'+b'+c')/3 = (ka' + kb' + kc')/3\; [\geq , \leq ]\; k \sqrt[3]{a' b' c'} = \sqrt[3]{ka' \cdot kb' \cdot kc'}$$ We can always take $k$ to give us $(ka' + kb' + kc') = 1$, then call $ka', kb', kc'$ by the new names $a,b,c$.

Oh, I see. Thanks for explaining. Now I have seen more ways to prove this than anyone in my class will probably ever want to see xD.