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?

  • 1
    $\begingroup$ Something is wrong here, since this is false when $x=1$ and $y=z=-0.5$: $$(1+(-0.5)+(-0.5))^3=0^3=0\not\geq 6.75=27(1)(-0.5)(-0.5)$$ $\endgroup$ – Zev Chonoles Jun 30 '11 at 7:21
  • $\begingroup$ @Zev: Thanks for the correction! $\endgroup$ – amy Jun 30 '11 at 7:33

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 $x=-1, y=-2, z=-3$.

  • 1
    $\begingroup$ 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. $\endgroup$ – Mark Bennet Jun 30 '11 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 $n-1$. $\endgroup$ – Beni Bogosel Jun 30 '11 at 21:55
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    $\begingroup$ I believe Cauchy's proof of AM-GM is something like this. $\endgroup$ – Aryabhata Jul 1 '11 at 15:59

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.

  • $\begingroup$ very nice idea! $\endgroup$ – the L Jul 1 '11 at 12:31
  • $\begingroup$ Nice idea, surprised that it actually works but it not that popular... $\endgroup$ – Srivatsan Sep 2 '11 at 15:22
  • $\begingroup$ Interesting indeed. $\endgroup$ – user1551 Sep 2 '11 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 '11 at 20:44
  • $\begingroup$ We want symmetry just to break it and then the proof follows :D $\endgroup$ – Ennar Jul 2 '16 at 11:13

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

  • 1
    $\begingroup$ This is better than AM$\ge$GM because the variables are not required to be all positive. $\endgroup$ – user1551 Sep 2 '11 at 17:40

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.

  • $\begingroup$ With the AM-GM inequality this is almost a one line proof. $\endgroup$ – user38268 Jun 30 '11 at 7:30

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.

  • $\begingroup$ I don't really see the difference of this vs directly quoting AM-GM. $\endgroup$ – Soarer Jun 30 '11 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. $\endgroup$ – Vladimir Sotirov Jun 30 '11 at 17:09

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) $$ (x+y+z)^3 = 27 x y z + (x^3 + y^3 + z^3 - 3 x y z) + 3 (z-y)^2 x + 3 (x-z)^2 y+ 3 (y-x)^2 z $$ All terms (the second one in brackets) on the RHS are positive, for the second term this is true again by AM-GM: $ (x^3 + y^3 + z^3)/3 \geq \sqrt[3]{x^3 y^3 z^3} = x y z $. 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.


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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