How prove this $\prod\limits_{cyc}(a^3+2b+\frac{2}{a^2+1})\ge 64$

Question

let $a,b,c>0$ and such $$abc\ge 1$$ show that $$\left(a^3+2b+\dfrac{2}{a^2+1}\right)\left(b^3+2c+\dfrac{2}{b^2+1}\right)\left(c^3+2a+\dfrac{2}{c^2+1}\right)\ge 64$$

my try: $$\sum_{cyc}\ln{\left(a^3+2b+\dfrac{2}{a^2+1}\right)}\ge 6\ln{2}$$

Then I can't,Thank you

By the Purkiss Principle ( from http://www.maa.org/programs/maa-awards/writing-awards/do-symmetric-problems-have-symmetric-solutions ) we have that both expressions are minimal for: $(a,b,c) = (1,1,1)$ . But then we have to prove (again and again) that a local minimum is a global minimum. Doing that with help of pictures seems not to be convincing. Then you can't? Neither can I :-( — Han de Bruijn, Jan 20 '14 at 16:09

score 2 · Answer 1 · answered Jan 18 '14 at 13:20

2

Hint: first prove that $a^3+\frac{2}{a^2+1} \geq 2a$

answered Jan 18 '14 at 13:20

user68061

3,827

Hint: having become familiar with the (tough) questions asked by china_math , I don't think it's homework. – Han de Bruijn Jan 20 '14 at 15:52

How prove this $\prod\limits_{cyc}(a^3+2b+\frac{2}{a^2+1})\ge 64$

1 Answers1