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
\[\sum_{\text{sym}} x_1^{a_1} x_2^{a_2} \ldots x_n^{a_n}\]
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 \( 7+3+1=5+4+2.\)
Statement of the Inequality
Muirhead's Inequality:
Suppose \( (a_i)\) majorizes \( (b_i).\) Then, for all nonnegative \( x_i,\)
\[\sum_{\text{sym}} x_1^{a_1}\ldots x_n^{a_n} \ge \sum_{\text{sym}} x_1^{b_1}\ldots x_n^{b_n}.\]
From the example in the previous section, if \(x,y,z\) are nonnegative real numbers, then
\[\begin{align} x^7y^3z + x^7yz^3 + x^3y^7z + x^3yz^7 + xy^7z^3 +xy^3z^7 \ge x^5y^4z^2 + x^5y^2z^4 + x^4y^5z^2 + x^4y^2z^5 + x^2y^5z^4 + x^2y^4z^5.\end{align}\]
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 \( M(a_1,\ldots,a_n) \ge M(b_1,\ldots,b_n).\)
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 \( x_1^1 x_2^0 x_3^0 \ldots x_n^0\) 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
\[\begin{align} (n-1)!(x_1 + x_2 + \cdots + x_n) &\ge n!\left(x_1^{1/n} x_2^{1/n} \ldots x_n^{1/n}\right) \\ \frac{x_1+x_2+\cdots+x_n}{n} &\ge (x_1x_2\ldots x_n)^{1/n}, \end{align}\]
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 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 \( 8.\) \(_\square\)
\[x = a+b+c, \qquad y = \frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab}\]
If \(a,b,c\) are positive real numbers, which is bigger, \(x\) or \(y?\)
\[x^6+5x^5y+10x^4y^2+kx^3y^3+10x^2y^4+5xy^5+y^6\ge 0\]
Find the absolute value of the smallest possible \(k\) such that the inequality above is true for all non-negative reals \(x\) and \( y \).
\(\)
Note: You may use the algebraic identities below.
- \((x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5\)
- \((x+y)^6=x^6+6 x^5 y+15 x^4 y^2+20 x^3 y^3+15 x^2 y^4+6 x y^5+y^6\)
In the \(xyz\)-plane above, let \(O=(0,0,0)\) be the point of origin, \(P=(a,b,c)\), and \(Q=(a+b, b+c, c+a)\), where \(a,b,c\) are non-zero real numbers.
What is the maximum value of the ratio \(\frac{OQ}{OP}?\)
\[S = \left \{\left(\frac{x+y-z}{x+y+z}\right)^2 + \left(\frac{x-y+z}{x+y+z}\right)^2 + \left(\frac{-x+y+z}{x+y+z}\right)^2: x,y,z \in \mathbb R^+ \right \}\]
Let \(S\) be a set defined as above, where \(x, y, z\) are positive real numbers.
If \(a = \text{sup } S \) \((\)supremum of \(S)\) and \(b = \text{inf } S \) \((\)infimum of \(S),\) compute \(\frac{a}{b}\).
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,
\[\begin{align} M(10,0,0) &\ge M\left(9\!\frac13,\frac13,\frac13\right) \\ 2!\left(x^{10}+y^{10}+z^{10}\right) &\ge 2!\left(x^9(xyz)^{1/3} + y^9(xyz)^{1/3} + z^9(xyz)^{1/3}\right) \\ x^{10}+y^{10}+z^{10} &\ge x^9+y^9+z^9.\ _\square \end{align}\]
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
\[\frac1{a^3(b+c)} + \frac1{b^3(c+a)} + \frac1{c^3(a+b)} \ge \frac32.\]
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. \(_\square\)