Inequality: $(x + y + z)^3 \geq 27 xyz$

Solution 1:

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

Solution 2:

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.

Solution 3:

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

Solution 4:

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.