Positive real numbers $a,b,c$ satisfy $$2a^3b+2b^3c+2c^3a=a^2b^2+b^2c^2+c^2a^2.$$Prove that $$2ab(a-b)^2+2bc(b-c)^2+2ca(c-a)^2 \geq(ab+bc+ca)^2.$$
Problem
Source: 239 2015 S P6
Tags: inequalities
30.01.2021 12:12
matinyousefi wrote: Positive real numbers $a,b,c$ satisfy $$2a^3b+2b^3c+2c^3a=a^2b^2+b^2c^2+c^2a^2.$$Prove that $$2ab(a-b)^2+2bc(b-c)^2+2ca(c-a)^2 \geq(ab+bc+ca)^2.$$
30.01.2021 15:15
Quantum_fluctuations, it should be $$\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc)\geq0,$$which is obvious.
30.01.2021 15:22
arqady wrote: Quantum_fluctuations, it should be $$\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc)\geq0,$$which is obvious. Does it have any nice solution without using Buffalo method, sir?
30.01.2021 23:47
I think BW here it's just nice enough. Also, the $uvw$'s technique helps.
31.01.2021 01:37
$$\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc) = \sum_{cyc}ab(2a-b-c)^2 \geq 0$$
31.01.2021 09:51
@above identical to your solution 3 years ago
02.02.2021 04:08
Gems98 wrote: $$\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc) = \sum_{cyc}ab(2a-b-c)^2 \geq 0$$ Would you like to elaborate on how did you see it was just $ab(2a-b-c)^2$?
17.07.2024 15:19
Gems98 wrote: $$\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc) = \sum_{cyc}ab(2a-b-c)^2 \geq 0$$ mind if u demonstrate, i actually can`t to it