MHB Prove the minimum is greater than or equal to 8

[anemone]

Hello MHB!

Let $a,\,b$ and $c$ be positive real such that $(a+2)(b+2)(c+2)=27$. Prove that $(a^2+1)(b^2+1)(c^2+1)\ge 8$.

I've recently encountered the above really hard and good inequality problem that I failed to prove, I've been trying it for days but ALL my attempts proved futile. I feel completely defeated, and hope to get some useful hints to solve it from MHB...

Thanks in advance for the help!

 

[kaliprasad]

Hello MHB!
I've recently encountered the above really hard and good inequality problem that I failed to prove, I've been trying it for days but ALL my attempts proved futile. I feel completely defeated, and hope to get some useful hints to solve it from MHB...

Thanks in advance for the help!

Because of cyclic symmetry we can see that extremum is ar a = b= c
this gives $(a+2)^3 = 27$ or a = 1 so a = b = c = 1

now $(a^2+1)(b^2+1)(c^2 +1) = 8$

taking a = b = 2 we get c = 4.75 and $(a^2+1)(b^2+1)(c^2+1) = 5 * 5 * (4.75^2+1) > 8$ and hence the result

 

[anemone]

Thanks kaliprasad...that advice does help!:)