Prove or disprove:

If $\max(a_1,a_2)\ge\max(b_1,b_2)$, then $$x^{a_1}y^{a_2}+x^{a_2}y^{a_1}\ge x^{b_1}y^{b_2}+x^{b_2}y^{b_1}$$

I can not understand it in the proof Muirhead's inequality


Muirhead's inequality.


Case 1. Let $b_1 \ge a_2$. Then $\max(a_1,a_2)\ge\max(a_1+a_2-b_1,b_1)$ and $\max(a_1+a_2-b_1,a_3)\ge\max(b_2,b_3)$. Then $$\sum_{sym}x^{a_1}y^{a_2}z^{a_3}=\sum_{cyc}z^{a^3}\left(x^{a_1}y^{a_2}+x^{a_2}y^{a_1}\right)\ge$$ $$\ge\sum_{cyc}z^{a^3}\left(x^{a_1+a_2-b_1}y^{b_1}+x^{b_1}y^{a_1+a_2-b_1}\right)\ge...$$

Q.: Why $$\sum_{cyc}z^{a^3}\left(x^{a_1}y^{a_2}+x^{a_2}y^{a_1}\right)\ge \sum_{cyc}z^{a^3}\left(x^{a_1+a_2-b_1}y^{b_1}+x^{b_1}y^{a_1+a_2-b_1}\right)?$$

  • 2
    $\begingroup$ This looks like it would be false. If $y = x$, then this reduces to $2x^{a_1 + a_2} \ge 2 x^{b_1 + b_2}$, which, if $x \ge 1$, is equivalent to $a_1 + a_2 \ge b_1 + b_2$. However, the condition $\max(a_1, a_2) \ge \max(b_1, b_2)$ is not sufficient to guarantee this. For a concrete counterexample, take $x = y = 2$ and $a_1 = 4, a_2 = 1, b_1 = b_2 = 3$. $\endgroup$
    – Shagnik
    Jul 25 '16 at 8:26
  • 1
    $\begingroup$ (I think for Muirhead's inequality you also require $a_1 + a_2 = b_1 + b_2$.) $\endgroup$
    – Shagnik
    Jul 25 '16 at 8:29

In http://www.mathsolympiad.org.nz/wp-content/uploads/2008/12/inequalities.pdf page 13 / No. 6 is your question answered.


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