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Edit: $a,b,c$ and $x,y,z$ are positive, real numbers.

Since $(a-b)^2 \geq 0~$, $a^2 + b^2 - 2ab\geq0~$ and $a^2 + b^2 \geq 2ab~$. Similarly, $a^2 + c^2 \geq 2ac~$ and $b^2 + c^2 \geq 2bc~$.

Adding these inequalities together, $2(a^2 +b^2 + c^2) \geq 2(ab + ac +bc)~$ and accordingly, $a^2 +b^2 + c^2 \geq ab + ac +bc~$

Multiplying both sides by $(a + b + c)$:

$(a^2 +b^2 + c^2)(a+b +c) \geq (ab + ac +bc)(a + b + c)~~$ and simplifying this, $ a^3 + b^3 + c^3 + \Sigma a^2 b \geq 3abc + \Sigma a^2 b $

Therefore, it follows that $a^3 + b^3 + c^3 \geq 3abc~$, and letting $a^3 = x~$, $b^3 = y~$, $c^3 = z~$: $x + y + z\geq 3\sqrt[3]{xyz}$

Cubing both sides, $(x + y + z)^3 \geq 27 xyz~$ which was to be proven.

I was wondering if there are alternative approaches to solve this problem (possibly using higher-level maths), and is my proof entirely correct?

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    $\begingroup$ Something is wrong here, since this is false when $x=1$ and $y=z=-0.5$: $$(1+(-0.5)+(-0.5))^3=0^3=0\not\geq 6.75=27(1)(-0.5)(-0.5)$$ $\endgroup$ Jun 30, 2011 at 7:21
  • $\begingroup$ @Zev: Thanks for the correction! $\endgroup$
    – astiara
    Jun 30, 2011 at 7:33

8 Answers 8

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I know a nice proof. It goes like this:

Let $x,y,z>0$. You know that $\frac{x+y}{2} \geq \sqrt{xy}$. This can be generalized for four numbers $$\frac{a+b+c+d}{4}=\frac{\frac{a+b}{2}+\frac{c+d}{2}}{2}\geq \sqrt[4]{abcd}.$$

Now pick $a=x,b=y,c=z,d=\sqrt[3]{xyz}$ and you'll get your inequality.

For $x,y,z$ not positive the inequality may not hold. Check $x=-1, y=-2, z=-3$.

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    $\begingroup$ I like this a lot. I remember when our lecturer put up a proof of AM/GM he did the $2^r$ case easily and there was some inelegant fudge for the general case. This way you can do $2^r$ and there is a nice (and rare) downward induction based on a generalisation of this observation. I haven't seen it done like this before. $\endgroup$ Jun 30, 2011 at 18:51
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    $\begingroup$ This type of argument can be applied to prove AM-GM inequality using induction in a different way, namely the inequality for $n$ implies the inequality for $2n$, and the inequality for $n$ implies the inequality for $n-1$. $\endgroup$ Jun 30, 2011 at 21:55
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    $\begingroup$ I believe Cauchy's proof of AM-GM is something like this. $\endgroup$
    – Aryabhata
    Jul 1, 2011 at 15:59
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My favorite technique for proving symmetric inequalities of positive numbers (particularly if you have a computer algebra package) is to note that if the inequality is symmetric, then w.l.o.g. we can assume the variables are in sorted order, then rewrite the inequality using the smallest variable and the consecutive differences, expand everything algebraically and note that all the coefficients are positive.

Using the example at hand

$(x + y + z)^3 - 27 x y z \ge 0$

assume w.l.o.g. $x\le y \le z$ and let $y=x+a$ and $z = x + a + b$, so

$\begin{align*}(x + y + z)^3 - 27 x y z &= (3x + 2a + b)^3 - 27 x (x+a)(x+a+b) \\ &= 9 a b x + 6 a b^2 + 9 x a^2 + 9 x b^2 + 12 b a^2 + b^3 + 8a^3\end{align*}$

which is greater than or equal to $0$ as all of $x$, $a$, and $b$ are.

This trick does not always work, but it works surprisingly often.

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  • $\begingroup$ very nice idea! $\endgroup$
    – the L
    Jul 1, 2011 at 12:31
  • $\begingroup$ Nice idea, surprised that it actually works but it not that popular... $\endgroup$
    – Srivatsan
    Sep 2, 2011 at 15:22
  • $\begingroup$ Interesting indeed. $\endgroup$
    – user1551
    Sep 2, 2011 at 17:46
  • $\begingroup$ IIRC it is a trick of Thomas Mautsch that was discussed on sci.math (see mathforum.org/kb/… ). There is also a nice long expository post by Dave Rusin somewhere around there, but I cannot find it. $\endgroup$
    – deinst
    Sep 2, 2011 at 20:44
  • $\begingroup$ We want symmetry just to break it and then the proof follows :D $\endgroup$
    – Ennar
    Jul 2, 2016 at 11:13
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An elementary approach, without $\text{AM} \ge \text{GM}$ is to use the identity

$$x^3 + y^3 + z^3 - 3xyz = (x+y+z)\left(\frac{(x-y)^2 + (y-z)^2 + (z-x)^2}{2}\right)$$

Thus $$\text{if } \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge 0\ \text{then } (a+b+c)^3 \ge 27abc$$

by setting $x = \sqrt[3]{a}$ etc

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    $\begingroup$ This is better than AM$\ge$GM because the variables are not required to be all positive. $\endgroup$
    – user1551
    Sep 2, 2011 at 17:40
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Look up "arithmetic-geometric mean inequality". Your proof is fine, if you assume the variables $\ge 0$, except that your notation $\Sigma a^2 b$ is nonstandard.

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  • $\begingroup$ With the AM-GM inequality this is almost a one line proof. $\endgroup$
    – user38268
    Jun 30, 2011 at 7:30
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The following uses what is perhaps the most practical formulation of the AM-GM inequalities, which can be found as Theorem 2.6a in Ivan Niven's excellent Maxima and Minima without Calculus.

Theorem 2.6a If $n$ positive functions have a fixed product, their sum is minimum if it can be arranged that the functions are equal. On the other hand, if $n$ positive functions have a fixed sum, their product is maximum if it can be arranged that the functions are equal.

By "positive" functions, Niven means functions that are positive on the domain we care about.

To see how this applies to the problem at hand, we see that it is enough to show that $(x+y+z)^3\geq 27k$ where $k$ is the product of $xyz$. Evidently, the theorem applies as the product is constant, so the minimum on the left-hand side is given by when $x=y=z=\sqrt[3]{k}$, and hence the minimum of the left-hand side is in fact $27k$ as desired.


The statement and (one) proof of the AM-GM inequalities can be found on page 21 of Niven's book, while the proof of Theorem 2.6a begins at the bottom of page 27.

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  • $\begingroup$ I don't really see the difference of this vs directly quoting AM-GM. $\endgroup$
    – user325
    Jun 30, 2011 at 16:55
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    $\begingroup$ Of course not: the two formulations are logically equivalent, and this particular question happens to not be particularly complicated. Consider maximizing something like $xy(72-3x-4y)$, which would be vaguely painful using calculus, or with slight massage can be rewritten as $1/12(3x)(4y)(72-3x-4y)$, where the individual terms now have constant sum, so maximum is when $3x=4y=72-3x-4y$, which evidently should equal $72/3=24$, so $x=8,y=6$. Personally, I never found the AM-GM particularly memorable or knew how to use it before reading of applications like this one in Niven's book. $\endgroup$ Jun 30, 2011 at 17:09
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Since the inequality is homogeneous we may WLOG assume that $xyz=1$ and we have to prove the inequality $(x+y+z)^3 \geq 27$ or $x+y+z \geq 3.$ But the last inequality we obtain immediately from $AM \geq GM$ under the assumption $xyz=1.$

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Here are two one-line-formula proofs:

  1. By AM-GM: $$ \frac{x+y+z}{3} \geq \sqrt[3]{x y z} $$ Now take the cubic value on both sides of this inequality.

  2. Use the following identity which also gives you the exact deviation in positive terms from $27 x y z$: (from which you can derive tighter bounds of the LHS) $$ (x+y+z)^3 = 27 x y z + 3 (z-y)^2 x + 3 (x-z)^2 y+ 3 (y-x)^2 z +\\ + \frac12 (x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2) $$ All terms on the RHS are positive, so you can take lower bounds of the LHS by any term on the RHS or weighted sum of terms on the RHS, with weights between 0 and 1.

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I was wondering if there are alternative approaches to solve this problem

Yes, there is.

and is my proof entirely correct?

I think so. I couldn't find anything wrong with it.


Now let's start with another proof.

Let $$ \{a,b,c,x,y,z \in \mathbb R_{\geq0} : a=x^{\frac{1}{3}},b=y^{\frac{1}{3}},c=z^{\frac{1}{3}}\} $$

We know that $$ \begin{equation} \begin{aligned} &a^2+b^2 \geq 2ab \\ \implies &(a+b)(a^2+b^2-ab) \geq ab(a+b) \\ \end{aligned} \end{equation} $$

$$ a^3+b^3 \geq ab(a+b) \label{eq1} \tag{1} $$

Utilizing this result across $a$, $b$ and $c$ and adding them, we get $$ 2(a^3+b^3+c^3) \geq a^2(b+c)+b^2(c+a)+c^2(a+b) \label{eq2} \tag{2} $$

Now, $$ a+b \geq 2\sqrt{ab} $$

Utilizing this result across $a$, $b$ and $c$ and multiplying them, we get $$ (a+b)(b+c)(c+a) \geq 8abc $$

On further simplification, we get $$ a^2(b+c)+b^2(c+a)+c^2(a+b) \geq 6abc \label{eq3} \tag{3} $$

From $\eqref{eq2}$ and $\eqref{eq3}$, it's clear that $$ 2(a^3+b^3+c^3) \geq a^2(b+c)+b^2(c+a)+c^2(a+b) \geq 6abc $$

Therefore, $$ \begin{aligned} &2(a^3+b^3+c^3) \geq 6abc \\ \implies &a^3+b^3+c^3 \geq 3abc \\ \implies &x+y+z \geq 3(xyz)^{\frac{1}{3}} \end{aligned} $$

Cubing on both sides, we get $$ (x+y+z)^3 \geq 27xyz $$

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