# 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.

## Preliminary definitions

Let \( a_1,\ldots,a_n\) be nonnegative real numbers. Let \( x_1,\ldots,x_n\) be variables. Then \[ \sum_{\text{sym}} x_1^{a_1} x_2^{a_2} \cdots x_n^{a_n} \] is the sum of \(n!\) terms of the form \( x_{\sigma(1)}^{a_1} x_{\sigma(2)}^{a_2} \cdots x_{\sigma(n)}^{a_n},\) 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 \( 7+3+1=5+4+2.\)

## Statement of the inequality

Muirhead's inequality:Suppose \( (a_i)\) majorizes \( (b_i).\) Then, for all nonnegative \( x_i,\) \[ \sum_{\text{sym}} x_1^{a_1}\cdots x_n^{a_n} \ge \sum_{\text{sym}} x_1^{b_1}\cdots x_n^{b_n}. \]

From the example in the previous section, if \(x,y,z\) are nonnegative real numbers, then \[ \begin{align} x^7y^3z + x^7yz^3 + x^3y^7z + x^3yz^7 + xy^7z^3 +xy^3z^7 \ge \phantom{aaaaaaaaaaaaaa} \\ x^5y^4z^2 + x^5y^2z^4 + x^4y^5z^2 + x^4y^2z^5 + x^2y^5z^4 + x^2y^4z^5. \end{align} \]

In the interest of more compact notation, write \[ M(a_1,a_2,\ldots,a_n) = \sum_{\text{sym}} x_1^{a_1}\cdots x_n^{a_n}. \] Then Muirhead's inequality says that if \( (a_i)\) majorizes \( (b_i),\) then \( M(a_1,\ldots,a_n) \ge M(b_1,\ldots,b_n).\)

## Applications

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 \( x_1^1 x_2^0 x_3^0 \cdots x_n^0\) is preserved by all the permutations that keep \( 1\) fixed and move \( 2\cdots n\) 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 \[ \begin{align} (n-1)!(x_1 + x_2 + \cdots + x_n) &\ge n!(x_1^{1/n} x_2^{1/n} \cdots x_n^{1/n}) \\ \frac{x_1+x_2+\cdots+x_n}{n} &\ge (x_1x_2\cdots x_n)^{1/n} \end{align} \] 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 the positive real numbers.

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 \( 8.\)

\[x^6+5x^5y+10x^4y^2+kx^3y^3+10x^2y^4+5xy^5+y^6\ge 0,\]

Find the absolute value of the smallest possible value for \(k\), such that the inequality above is true for all non-negative reals \(x\) and \( y \).

**Details and Assumptions**

- You may use the following algebraic identities.
\((x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5\)

\((x+y)^6=x^6+6 x^5 y+15 x^4 y^2+20 x^3 y^3+15 x^2 y^4+6 x y^5+y^6\)

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 \( x^{10}+y^{10}+z^{10} \ge x^9+y^9+z^9.\)

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: \[ \begin{align} M(10,0,0) &\ge M\left(9\!\frac13,\frac13,\frac13\right) \\ 2!(x^{10}+y^{10}+z^{10}) &\ge 2!(x^9(xyz)^{1/3} + y^9(xyz)^{1/3} + z^9(xyz)^{1/3}) \\ x^{10}+y^{10}+z^{10} &\ge x^9+y^9+z^9. \end{align} \]

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

(IMO 1995) For \(a,b,c>0, abc=1,\) prove that \[ \frac1{a^3(b+c)} + \frac1{b^3(c+a)} + \frac1{c^3(a+b)} \ge \frac32. \]

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 \[ M(4,3,1) + \frac12 M(4,4,0) + \frac12 M(3,3,2) \ge \frac32(2abc+M(2,1,0)) = 3abc + \frac32 M(2,1,0). \] 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 \((abc)^{8/3} = \frac16 M(8/3,8/3,8/3).\) So: \[ M(4,3,1) + \frac12 M(4,4,0) + \frac12 M(3,3,2) \ge \frac12 M(8/3,8/3,8/3)+\frac32 M(11/3,8/3,5/3). \] Now \(4,3,1\) and \(4,4,0\) both majorize \( 11/3,8/3,5/3,\) so \[ M(4,3,1)+\frac12M(4,4,0) \ge M(11/3,8/3,5/3)+\frac12 M(11/3,8/3,5/3) = \frac32 M(11/3,8/3,5/3), \] and \( 3,3,2\) majorizes \(8/3,8/3,8/3,\) so \[ \frac12 M(3,3,2) \ge \frac12 M(8/3,8/3,8/3), \] and adding these inequalities gives the result.

**Cite as:**Muirhead Inequality.

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