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.

  • 6
    $\begingroup$ The value is non-negative and zero for $a=b=c=-1$. $\endgroup$
    – Dirk
    Mar 2, 2012 at 8:30
  • 4
    $\begingroup$ I have tagged it as algebra-precalculus, as I suppose you don't want a calculus answer. Also, are there any restrictions on $a,b,c$? Like positive reals and sum is constant or something like that? $\endgroup$
    – Aryabhata
    Mar 2, 2012 at 8:32

1 Answer 1


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?

  • $\begingroup$ @vikiiii: You are welcome. $\endgroup$
    – Aryabhata
    Mar 2, 2012 at 9:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .