MHB Prove the minimum is greater than or equal to 8

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The inequality problem involves proving that for positive reals \(a\), \(b\), and \(c\) satisfying \((a+2)(b+2)(c+2)=27\), it holds that \((a^2+1)(b^2+1)(c^2+1) \ge 8\). A key insight is that due to cyclic symmetry, the extremum occurs when \(a = b = c = 1\), yielding \((1^2+1)^3 = 8\). Testing other values, such as \(a = b = 2\) and \(c = 4.75\), shows that the product exceeds 8, confirming the inequality. The discussion highlights the challenge of the problem and the eventual realization that symmetry simplifies the proof. The conclusion is that the minimum value of the expression is indeed greater than or equal to 8.
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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.:mad: I feel completely defeated, and hope to get some useful hints to solve it from MHB...

Thanks in advance for the help!
 
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anemone said:
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.:mad: 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
 
Thanks kaliprasad...that advice does help!:)
 
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