Muirhead Inequality

Muirhead's inequality is a generalization of the AM-GM inequality. Like the AM-GM inequality, it involves a comparison of symmetric sums of monomials involving several variables. It is often useful in proofs involving inequalities.

Let $a_1,\ldots,a_n$ be nonnegative real numbers. Let $x_1,\ldots,x_n$ be variables. Then

$$\sum_{\text{sym}} x_1^{a_1} x_2^{a_2} \ldots x_n^{a_n}$$

is the sum of $n!$ terms of the form $x_{\sigma(1)}^{a_1} x_{\sigma(2)}^{a_2} \ldots x_{\sigma(n)}^{a_n},$ where $\sigma$ runs over the permutations of $\{1, \ldots,n\}.$

$$\sum_{\text{sym}} xy^3z^2 = xy^3z^2+xy^2z^3 + x^2y^3z + x^2yz^3 + x^3y^2z + x^3yz^2$$

Now suppose $a_1,\ldots, a_n$ and $b_1,\ldots,b_n$ are two sequences of nonnegative real numbers satisfying the following conditions:

(1) $a_1\ge a_2\ge \cdots \ge a_n$ and $b_1\ge b_2\ge \cdots \ge b_n$
(2) $a_1 \ge b_1,$ $a_1+a_2 \ge b_1+b_2,$ $\ldots,$ $a_1+a_2+\cdots+a_{n-1} \ge b_1+b_2+\cdots+b_{n-1}$
(3) $a_1+a_2+\cdots+a_n = b_1+b_2+\cdots + b_n.$

Then the sequence $(a_i)$ is said to majorize the sequence $(b_i).$

$7,3,1$ majorizes $5,4,2,$ because $7 \ge 5,$ $7+3 \ge 5+4,$ and $7+3+1=5+4+2.$

Muirhead's Inequality:

Suppose $(a_i)$ majorizes $(b_i).$ Then, for all nonnegative $x_i,$

$$\sum_{\text{sym}} x_1^{a_1}\ldots x_n^{a_n} \ge \sum_{\text{sym}} x_1^{b_1}\ldots x_n^{b_n}.$$

From the example in the previous section, if $x,y,z$ are nonnegative real numbers, then

$$\begin{aligned} x^7y^3z + x^7yz^3 + x^3y^7z + x^3yz^7 + xy^7z^3 +xy^3z^7 \ge x^5y^4z^2 + x^5y^2z^4 + x^4y^5z^2 + x^4y^2z^5 + x^2y^5z^4 + x^2y^4z^5.\end{aligned}$$

In the interest of more compact notation, write

$$M(a_1,a_2,\ldots,a_n) = \sum_{\text{sym}} x_1^{a_1}\ldots x_n^{a_n}.$$

Then Muirhead's inequality says that if $(a_i)$ majorizes $(b_i),$ then $M(a_1,\ldots,a_n) \ge M(b_1,\ldots,b_n).$

Since $(1,0,0,\ldots,0)$ majorizes $\left( \frac1{n}, \frac1{n}, \ldots, \frac1{n}\right),$ $M(1,0,0,\ldots,0) \ge M\left( \frac1{n},\frac1{n},\ldots,\frac1{n}\right)$ by Muirhead's inequality. In the symmetric sum for $M(1,0,0,\ldots,0),$ there are $(n-1)!$ permutations of the variables that give each term. For instance, the term $x_1^1 x_2^0 x_3^0 \ldots x_n^0$ is preserved by all the permutations that keep $1$ fixed and move $2\ldots n$ around. In the symmetric sum for $M\left( \frac1{n},\frac1{n},\ldots,\frac1{n}\right),$ all of the $n!$ permutations give the same monomial.

So Muirhead's inequality becomes

$$\begin{aligned} (n-1)!(x_1 + x_2 + \cdots + x_n) &\ge n!\left(x_1^{1/n} x_2^{1/n} \ldots x_n^{1/n}\right) \\ \frac{x_1+x_2+\cdots+x_n}{n} &\ge (x_1x_2\ldots x_n)^{1/n}, \end{aligned}$$

which is the AM-GM inequality. So this is a special case of Muirhead's inequality.

Find the minimum value of

$$\frac{(a+b)(b+c)(c+a)}{abc}$$

as $a,b,c$ range over positive real numbers.

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Multiplying out the numerator gives $2abc + M(2,1,0).$ Note that $M(1,1,1) = 6abc,$ so the expression is

$$2+\frac{M(2,1,0)}{abc} = 2+\frac{6 M(2,1,0)}{M(1,1,1)},$$

and $M(2,1,0) \ge M(1,1,1)$ by Muirhead's inequality, so the expression is $\ge 2+6 = 8.$ But note that if $a=b=c,$ then the expression evaluates to $8,$ so the answer is $8.$ $_\square$

Muirhead's inequality can sometimes be extended to prove non-homogeneous inequalities:

Suppose $x,y,z$ are positive real numbers with $xyz=1.$ Show that $x^{10}+y^{10}+z^{10} \ge x^9+y^9+z^9.$

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The trick is to multiply $x^9+y^9+z^9$ by $(xyz)^{1/3}=1,$ which makes the degrees of both sides equal to each other. That is,

$$\begin{aligned} M(10,0,0) &\ge M\left(9\!\frac13,\frac13,\frac13\right) \\ 2!\left(x^{10}+y^{10}+z^{10}\right) &\ge 2!\left(x^9(xyz)^{1/3} + y^9(xyz)^{1/3} + z^9(xyz)^{1/3}\right) \\ x^{10}+y^{10}+z^{10} &\ge x^9+y^9+z^9.\ _\square \end{aligned}$$

This last example is quite a bit harder, but illustrates the power of the technique.

[IMO 1995] For $a,b,c>0$ with $abc=1,$ prove that

$$\frac1{a^3(b+c)} + \frac1{b^3(c+a)} + \frac1{c^3(a+b)} \ge \frac32.$$

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Multiply by the product of denominators:

$$b^3c^3(c+a)(a+b) + a^3c^3(b+c)(a+b) + a^3b^3(b+c)(c+a) \ge \frac32 a^3b^3c^3(b+c)(c+a)(a+b).$$

Using $a^3b^3c^3=1$ on the right side, this turns into

$$M(4,3,1) + \frac12 M(4,4,0) + \frac12 M(3,3,2) \ge \frac32\big(2abc+M(2,1,0)\big) = 3abc + \frac32 M(2,1,0).$$

Now the left side has degree $8,$ and the right side has degree $3.$ To get the degrees on both sides to match up, we can multiply the right side by $(abc)^{5/3}=1.$ Note that $(abc)^{8/3} = \frac16 M\left(\frac83,\frac83,\frac83\right).$ So,

$$M(4,3,1) + \frac12 M(4,4,0) + \frac12 M(3,3,2) \ge \frac12 M\left(\frac83,\frac83,\frac83\right)+\frac32 M\left(\frac{11}3,\frac83,\frac53\right).$$

Now $4,3,1$ and $4,4,0$ both majorize $\frac{11}3,\frac83,\frac53,$ so

$$M(4,3,1)+\frac12M(4,4,0) \ge M\left(\frac{11}3,\frac83,\frac53\right)+\frac12 M\left(\frac{11}3,\frac83,\frac53\right) = \frac32 M\left(\frac{11}3,\frac83,\frac53\right),$$

and $3,3,2$ majorizes $\frac83,\frac83,\frac83,$ so

$$\frac12 M(3,3,2) \ge \frac12 M\left(\frac83,\frac83,\frac83\right),$$

and adding these inequalities gives the result. $_\square$