# Prove or disprove $x^{a_1}y^{a_2}+x^{a_2}y^{a_1}\ge x^{b_1}y^{b_2}+x^{b_2}y^{b_1}$

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

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 $$\sum_{cyc}z^{a^3}\left(x^{a_1}y^{a_2}+x^{a_2}y^{a_1}\right)\ge \sum_{cyc}z^{a^3}\left(x^{a_1+a_2-b_1}y^{b_1}+x^{b_1}y^{a_1+a_2-b_1}\right)?$$
• 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 $a_1 = 4, a_2 = 1, b_1 = b_2 = 3$. – Shagnik Jul 25 '16 at 8:26
• (I think for Muirhead's inequality you also require $a_1 + a_2 = b_1 + b_2$.) – Shagnik Jul 25 '16 at 8:29