Proving Cauchy inequality involving four expression

Question

Show that $$(a^2 + b^2 + c^2) (a^2b^2 +b^2c^2 +c^2a^2) \geq (a^2b + b^2c + c^2a)(ab^2 + bc^2 + ca^2)$$ i should prove this inequality by making it a Cauchy form inequality(as teacher stated).

my problem : this inequality is involving 4 parentheses but in other question i proved previously there was one or two in left side and one in right side.

things i have done so far:

$(a^2 + b^2 + c^2) (a^2b^2 +b^2c^2 +c^2a^2) \geq (a^2b + b^2c + c^2a)^2$

Answer 1

You say you have proved the inequality $$(a^2 + b^2 + c^2)(a^2b^2 + b^2c^2 + c^2a^2) \geq (a^2b + b^2c + c^2a)^2.$$ You are almost there! The square root of this equation tells us $$(a^2 + b^2 + c^2)^{1/2}(a^2b^2 + b^2a^2 + c^2a^2)^{1/2} \geq a^2b + b^2c + c^2a.$$

Now, you should be able to use identical arguments to prove that $$(a^2 + b^2 + c^2)^{1/2}(a^2b^2 + b^2a^2 + c^2a^2)^{1/2} \geq ab^2 + bc^c + ca^2.$$ Now what happens when you multiply the last two inequalities together?

Answer 2

Set $$P = (a^2 + b^2 + c^2) (a^2b^2 +b^2c^2 +c^2a^2) - (a^2b + b^2c + c^2a)(ab^2 + bc^2 + ca^2).$$ We have $$P = \frac14\sum c^2\left[a^2+b^2+c^2+(a+b-c)^2\right](a-b)^2 \geqslant 0.$$

Answer 3

It's $$\sum_{cyc}(a^4b^2+a^4c^2-a^3b^3-a^4bc)\geq0,$$ which is just Muirhead: $$\sum_{cyc}(a^4b^2+a^4c^2-2a^3b^3)+\sum_{cyc}(a^4b^2+a^4c^2-2a^4bc)\geq0,$$ which is true because $(4,2,0)\succ(3,3,0)$ and $(4,2,0)\succ(4,1,1)$.