<s>[INST] Finding the minimum value of $\left(\frac{a + 1}{a}\right)^2 + \left(\frac{b + 1}{b}\right)^2 + \left(\frac{c + 1}{c}\right)^2  $ Find the minimum value of 
$$\left(\frac{a + 1}{a}\right)^2 + \left(\frac{b + 1}{b}\right)^2 + \left(\frac{c + 1}{c}\right)^2  $$
I tried to expand it and break it into individual terms and use $\text{A.M} \ge \text{G.M}$. inequality. But in this case i stuck where i need to find the value of abc.
 [/INST] Assuming $a,b,c$ are positive reals, there is no minimum!
Consider the individual term: $$\left(\frac{a + 1}{a}\right)^2 = \left(1 + \frac{1}{a}\right)^2$$
For positive real $a$ this is strictly greater than $1$, but as $a \to \infty$, this can be made as close to $1$ as we want.
Thus the total sum is always strictly $\gt 3$, and can be made as close to $3$ as we want.
So there is no minimum (but there is an infimum).
If you allow $a,b,c$ to be negative, then $a=b=c=-1$ gives the minimum value of $0$, as pointed out in the comments.
Perhaps you are missing some condition? Like $a,b,c$ positive reals and $abc = 1$ or something like that?
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