Choosing suitable weights for AM-GM

Question

Consider the following example:

Prove that for all positive reals $a,b,c,d.$ $$a^4b+b^4c+c^4d+d^4a\ge abcd(a+b+c+d) $$ Solution: We apply AM-GM$$\frac{23a^4b+7b^4c+11c^4d+10d^4a}{51}\ge \sqrt[51]{a^{102}b^{51}c^{51}d^{51}}=a^2bcd $$ The solution follows easly from here. then OP states, one way to choose suitable weights is to suppose arbitrary ones $x_1,x_2,x_3,x_4$ for $a^4b,b^4c,c^4d,d^4a$ respectivly and solve the system of equations:$$x_1+x_2+x_3+x_4=1$$$$4x_1+x_2=2$$$$4x_2+x_3=1$$$$4x_3+x_4=1$$ The first equation is clear since the weights sum has to be $1$, the general case should be (correct me if I'm wrong) $$x_1a^4b+x_2b^4c+x_3c^4d+x_4d^4a\ge (a^{4x_1}b^{x_1})(b^{4x_2}c^{x_2})(c^{4x_3}d^{x_3})(d^{4x_4}a^{x_4})$$ thus the other 3 aren't very clear, why does he choose $4x_1+x_2=2,\:\: 4x_2+x_3=1, \:\: 4x_3+x_4=1$, isn't it supposed to be$$4x_1+x_4=2$$$$4x_2+x_1=1$$$$4x_3+x_2=1$$$$4x_4+x_3=1$$ (I guess we can also ignore one of the equations since we got 4 variables so we don't need 5 equations to solve the system) Am I wrong or did I do this correctly?

Answer 1

You did that correctly. Also, as you can see, the five equations which are arising are not independent, adding last 4 equations gives the first one.

Hope it helps:)

Answer 2

Also, you can use C-S and Muirhead.

Indeed, by C-S $$a^4b+b^4c+c^4d+d^4a=abcd\sum_{cyc}\frac{a^3}{cd}=abcd\sum_{cyc}\frac{a^4}{acd}\geq\frac{abcd\left(\sum\limits_{cyc}a^2\right)^2}{\sum\limits_{cyc}acd}.$$ Id est, it's enough to prove that $$(a^2+b^2+c^2+d^2)^2\geq(a+b+c+d)(abc+abd+acd+bcd)$$ or $$\sum_{cyc}a^4+\frac{1}{2}\sum_{sym}a^2b^2\geq\sum_{cyc}a^2(bc+bd+cd)+4abcd,$$ which is true by Muirhead because $(4,0,0,0)\succ(1,1,1,1)$ and $(2,2,0,0)\succ(2,1,1,0).$