## [Soln] Only AM-GM

Using AM-GM (Arithmetic Mean-Geometric Mean Inequality), prove that for $x, y, z \geq 0$,

$x^5yz + xy^5z + xyz^5 \geq x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3.$

The inequality is not a difficult one (in fact, it is an obvious corollary of Murihead’s inequality), but I just wanted to illustrate a technique of using AM-GM.

The AM-GM can’t be directly applied in the following standard ways because the terms on the RHS don’t work out:

$x^5yz + xy^5z \geq 2 \sqrt{x^6y^6z^2} = 2x^3y^3z,$
$x^5yz + xy^5z + zyz^5 \geq \sqrt[3]{x^7y^7z^7} = (xyz)^{7/3}.$

The trick here is that terms can be repeated multiple times in the AM-GM inequality. For example, instead of directly applying the AM-GM like this:

$3x + y \geq 2\sqrt{3xy},$

We could do this to get a different lower bound:

$3x + y = x + x + x + y \geq 4 \sqrt[4]{x^3y}.$

Applying this to our problem: we need to repeat the terms on the LHS an appropriate number of times such that the exponents of $x,y,z$ work out to be the exponents on the RHS. If we let the 3 terms on the LHS be repeated $a,b,c$ times respectively, we have

$ax^5yz + bxy^5z + cxyz^5 \geq (a+b+c) \sqrt[a+b+c]{x^{5a + b+ c}y^{a+5b+c}z^{a+b+5c}}.$

To make the RHS of the inequality above match the terms in the RHS of the original inequality, we need:

$\begin{cases} 5a + b + c &= 3(a+b+c), \\ a + 5b + c &= 2(a+b+c), \\ a + b + 5c &= 2(a+b+c). \end{cases}$

These can be solved through the usual method of elimination to give the solution $(a,b,c) = (2, 1, 1)$. (Note: The solution could have been guessed by some trial and error, but the above is a general method that doesn’t require any guessing.) Hence,

$x^5yz + x^5yz + xy^5z + xyz^5 \geq 4x^3y^2z^2.$

Similarly we find that

\begin{aligned} x^5yz + xy^5z + xy^5z + xyz^5 &\geq 4x^2y^3z^2, \\ x^5yz + xy^5z + xyz^5 + xyz^5 &\geq 4x^2y^2z^3. \end{aligned}

Adding up the last 3 inequalities and dividing by 4, we obtain the given inequality. Done!