## Inequalities of Schur and Muirhead Definition 1 We will consider the special cases of the functio \( F \), i.e. when \( F(a_1, \dots, a_n) = a_1^{\alpha_1}\cdot \cdots \cdot a_n^{\alpha_n} \), \( \alpha_i\geq 0 \). If \( (\alpha) \) is an array of exponents and \( F(a_1, \dots, a_n)=a_1^{\alpha_1}\cdot \cdots \cdot a_n^{\alpha_n} \) we will use \( T[\alpha_1, \dots,\alpha_n] \) instead of \( \sum! F(a_1, \dots, a_n) \), if it is clear what is the sequence \( (a) \). Example 1 \( T[1, 0, \dots, 0] = (n-1)!\cdot (a_1+a_2+\cdots + a_n) \), and \( T[\frac1n, \frac1n, \dots, \frac1n]= n! \cdot \sqrt[n]{a_1 \cdot \cdots \cdot a_n}. \) The AM-GM inequality is now expressed as: \[ T[1, 0, \dots, 0] \geq T\left[\frac1n, \dots, \frac1n\right].\] Theorem 1 (Schur) Let \( (x, y, z) \) be the sequence of positive reals for which we are proving (\ref{Schur}). Using some elementary algebra we get \begin{eqnarray*} && \frac12T[\alpha+2\beta, 0, 0] + \frac12T[\alpha, \beta, \beta] - T[\alpha+\beta, \beta, 0]\\ &=& x^{\alpha}(x^{\beta}-y^{\beta})(x^{\beta}-z^{\beta}) + y^{\alpha}(y^{\beta}-x^{\beta})(y^{\beta}-z^{\beta}) + z^{\alpha}(z^{\beta}-x^{\beta})(z^{\beta}-y^{\beta}). \end{eqnarray*} Without loss of generality we may assume that \( x \geq y \geq z \). Then in the last expression only the second summand may be negative. If \( \alpha\geq0 \) then the sum of the first two summands is \( \geq 0 \) because \[ x^{\alpha}(x^{\beta}-y^{\beta})(x^{\beta}-z^{\beta}) \geq x^{\alpha}(x^{\beta}-y^{\beta})(y^{\beta}-z^{\beta}) \geq y^{\alpha}(x^{\beta}-y^{\beta})(y^{\beta}-z^{\beta}) =- y^{\alpha}(x^{\beta}-y^{\beta})(y^{\beta}-z^{\beta}).\] Similarly for \( \alpha< 0 \) the sum of the last two terms is \( \geq 0 \). Example 2 If we set \( \alpha=\beta=1 \), we get \[ x^3+y^3 + z^3 +3xyz \geq x^2y+xy^2+ y^2z+yz^2+ z^2x+zx^2.\] Definition 2 We say that the array \( (\alpha) \) majorizes array \( (\alpha^{\prime}) \), and we write that in the following way \( (\alpha^{\prime})\prec (\alpha), \) if we can arrange the elements of arrays \( (\alpha) \) and \( (\alpha^{\prime}) \) in such a way that the following three conditions are satisfied: -
**(i)**\( \alpha_1^{\prime}+ \alpha_2^{\prime}+ \cdots + \alpha_n^{\prime} = \alpha_1 + \alpha_2 + \cdots + \alpha_n \); **(ii)**\( \alpha_1^{\prime}\geq \alpha_2^{\prime}\geq \cdots \geq \alpha_n^{\prime} \) i \( \alpha_1\geq \alpha_2 \geq \cdots \geq \alpha_n \).**(iii)**\( \alpha_1^{\prime}+ \alpha_2^{\prime}+ \cdots + \alpha_{\nu}^{\prime} \leq \alpha_1 + \alpha_2 + \cdots + \alpha_{\nu} \), for all \( 1\leq \nu < n \).
Clearly, \( (\alpha)\prec (\alpha) \). Theorem 2 (Muirhead) First, we prove the necessity of the condition. Setting that all elements of the array \( a \) are equal to \( x \), we get that \[ x^{\sum \alpha_i^{\prime}} \leq x^{\sum \alpha_i}.\] This can be satisfied for both large and small \( x \)s only if the condition \( 1 \) from the definition is satisfied. Now we put \( a_1= \cdots, a_{\nu} =x \) and \( a_{\nu+1}=\cdots = a_n=1 \). Comparing the highest powers of \( x \) in expressions \( T[\alpha] \) and \( T[\alpha^{\prime}] \), knowing that for sufficiently large \( x \) we must have \( T[\alpha^{\prime}] \leq T[\alpha] \), we conclude that \( \alpha_1^{\prime} + \cdots + \alpha_{\nu}^{\prime} \leq \alpha_1+ \cdots + \alpha_{\nu} \). Now we will proof the sufficiency of the condition. The statement will follow from the following two lemmas. We will define one linear operation \( L \) on the set of the exponents \( (\alpha) \). Suppose that \( \alpha_k \) and \( \alpha_l \) are two different exponents of \( (\alpha) \) such that \( \alpha_k > \alpha_l \). We can write \[ \alpha_k=\rho+\tau, \hspace{2mm} \alpha_l=\rho-\tau \hspace{2mm} (0< \tau \leq \rho).\] If \( 0\leq \sigma < \tau \leq \rho \), define the array \( (\alpha^{\prime})= L(\alpha) \) in the following way: \[ \left\{\begin{array}{l} \alpha_k^{\prime} = \rho+ \sigma = \frac{\tau+\sigma}{2\tau}\alpha_k + \frac{\tau-\sigma}{2\tau}\alpha_l,\\ \alpha_l^{\prime}=\rho-\sigma = \frac{\tau-\sigma}{2\tau}\alpha_k+ \frac{\tau+\sigma}{2\tau}\alpha_l,\\ \alpha_{\nu}^{\prime} = \alpha_{\nu}, \hspace{2mm} (\nu\neq k, \nu \neq l). \end{array} \right.\] The definition of this mapping doesn’t require that some of the arrays \( (\alpha) \) and \( (\alpha^{\prime}) \) is in non-decreasing order.
Example 1 (continued) Problem 1 After multiplying both left and right-hand side of the required inequality with \( abc(a^3+b^3 + abc)(b^3+ c^3+ abc)(c^3+a^3+abc) \) we get that the original inequality is equivalent to \[ \begin{array}{c} \frac32T[4,4,1] + 2T[5,2,2] + \frac12T[7,1,1]+\frac12T[3,3,3] \leq\\ \leq \frac12T[3,3,3]+ T[6,3,0] + \frac32 T[4,4,1] + \frac12T[7,1,1] + T[5,2,2]\end{array}\] which is true becaues Muirhead’s theorem imply that \( T[5,2,2] \leq T[6,3,0] \). The equality holds if and only if \( a=b=c \). Problem 2 The expressions have to be homogenous in order to apply the Muirhead’s theorem. First we divide both left and right-hand side by \( (abc)^{\frac43}=1 \) and after that we multiply both sides by \( a^3b^3c^3(a+b)(b+c)(c+a)(abc)^{\frac43} \). The inequality becomes equivalent to \[ 2T\left[\frac{16}3, \frac{13}3, \frac73\right]+ T\left[\frac{16}3, \frac{16}3, \frac43\right] + T\left[\frac{13}3, \frac{13}3, \frac{10}3\right] \geq 3T[5,4,3]+T[4,4,4].\] The last inequality follows by adding the following three which are immediate consequences of the Muirhead’s theorem: \[ \begin{array}{ll} 1.& 2T\left[\frac{16}3, \frac{13}3, \frac73\right]\geq 2T[5,4,3],\\ 2.& T\left[\frac{16}3, \frac{16}3, \frac43\right] \geq T[5,4,3], \\ 3.& T\left[\frac{13}3, \frac{13}3, \frac{10}3\right] \geq T[4,4,4].\end{array}\] The equality holds if and only if \( a=b=c=1 \). Problem 3 The left-hand side can be easily transformed into \( \frac{a^3(b+c)}{b^3+c^3}+ \frac{b^3(c+a)}{c^3+a^3}+ \frac{c^3(a+b)}{a^3+b^3} \). We now multiply both sides by \( (a+b+c)(a^3+b^3)(b^3+c^3)(c^3+a^3). \) After some algebra the left-hand side becomes \[ \begin{array}{ll}L=& T[9,2,0]+T[10,1,0] + T[9,1,1] + T[5,3,3] + 2T[4,4,3]\\ & + T[6,5,0]+ 2T[6,4,1]+ T[6,3,2]+ T[7,4,0] + T[7,3,1], \end{array}\] while the right-hand side transforms into \[ D = 3(T[4,4,3]+ T[7,4,0]+ T[6,4,1]+ T[7,3,1]).\] According to Muirhead’s theorem we have: \[ \begin{array}{ll} 1.& T[9,2,0]\geq T[7,4,0],\\ 2.& T[10,1,0] \geq T[7,4,0], \\ 3.& T[6,5,0] \geq T[6,4,1],\\ 4.& T[6,3,2]\geq T[4,4,3].\end{array}\] The Schur’s inequality gives us \( T[4, 2, 2]+ T[8,0,0] \geq 2T[6,2,0] \). After multiplying by \( abc \), we get: \[ \begin{array}{ll} 5.& T[5,3,3] + T[9,1,1]\geq T[7,3,1]. \end{array}\] Adding up \( 1, 2, 3, 4 \), \( 5 \), and adding \( 2T[4,4,3]+T[7,4,0]+2T[6,4,1]+ T[7,3,1] \) to both sides we get \( L\geq D \). The equality holds if and only if \( a=b=c \). Problem 4 (IMO 2005) Multiplying the both sides with the common denominator we get \[ T_{5,5,5}+4T_{7,5,0}+T_{5,2,2}+T_{9,0,0}\geq T_{5,5,2}+T_{6,0,0}+2T_{5,4,0}+2T_{4,2,0}+T_{2,2,2}.\] By Schur’s and Muirhead’s inequalities we have that \( T_{9,0,0}+T_{5,2,2}\geq 2T_{7,2,0}\geq 2T_{7,1,1} \). Since \( xyz\geq 1 \) we have that \( T_{7,1,1}\geq T_{6,0,0} \). Therefore \[ T_{9,0,0}+T_{5,2,2}\geq 2T_{6,0,0}\geq T_{6,0,0}+T_{4,2,0}. \] Moreover, Muirhead’s inequality combined with \( xyz\geq 1 \) gives us \( T_{7,5,0}\geq T_{5,5,2} \), \( 2T_{7,5,0}\geq 2T_{6,5,1}\geq 2T_{5,4,0} \), \( T_{7,5,0}\geq T_{6,4,2}\geq T_{4,2,0} \), and \( T_{5,5,5}\geq T_{2,2,2} \). Adding these four inequalities to (1) yields the desired result. Problem 5 After multiplying everything out the inequality becomes \[ T[2,1,0]\leq \frac12T[3,0,0]+\frac12T[1,1,1]. \]This follows immediately from the Schur’s inequality. Equality holds if and only if \( a=b=c \). Another way to do this is to use the substitution \( b+c-a=x \), \( c+a-b=y \), \( a+b-c=z \). After noticing that at most one of the factors can be negative (in which case the problem is trivial), we can multiply everything out and get some simpler expression which can be handled by Muirhead’s inequality. |

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