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Let $a, b, c \in \mathbb{R}^+$ such that $a^2 + b^2 + c^2 = 3$. Find the minimum value of $P = \dfrac{1}{(a-b)^2} + \dfrac{1}{(b-c)^2} + \dfrac{1}{(c-a)^2}$?

In fact, we have that $P \geq \dfrac{4}{ab +bc +ca}$. However, the equality can not occur in this exercise because $a >0, b > 0$ and $c> 0$. Would you please send me a hint for this problem? Thank you so much!

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    $\begingroup$ How about AM HM inequality? $\endgroup$ – lab bhattacharjee Jul 12 '18 at 4:17
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    $\begingroup$ The minimum will be as close as you want to get to $\frac{1}{6} \left(5 \sqrt{5}+11\right)$ with that value being the minimum only if you allow one of $a$, $b$, or $c$ equaling zero. $\endgroup$ – JimB Jul 12 '18 at 4:20
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    $\begingroup$ There is no minimum. The infimum is $\frac16(11+5\sqrt5)$, when one variable tends to zero and the others tend to the roots of a high order polynomial. Where did you get this problem - this doesn't seem one that you get to solve without CAS. $\endgroup$ – Macavity Jul 12 '18 at 4:20
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Let $D = \{ (a,b,c) \in (0,\infty)^3 : a^2 + b^2 + c^2 = 3 \}$ and $\bar{D}$ be its closure. Instead of looking for a minimum of $P = \frac{1}{(a-b)^2} + \frac{1}{(b-c)^2} + \frac{1}{(c-a)^2}$ over $D$, we need to look at the minumum over the larger $\bar{D}$. This is because $P$ doesn't achieve any local minimum over $D$.

Assume the contrary, let's say $P$ achieve a local minimum at $(a,b,c) \in D$. WOLOG, we can assume $0 < a < b < c$. For this to happen, the methods of Lagrange multipliers tell us there is a $\lambda \in \mathbb{R}$ such that $$ \begin{align} \frac{1}{(a-b)^3} + \frac{1}{(a-c)^3} &= \lambda a\\ \frac{1}{(b-a)^3} + \frac{1}{(b-c)^3} &= \lambda b\\ \frac{1}{(c-a)^3} + \frac{1}{(c-b)^3} &= \lambda c\\ \end{align} $$ Summing these 3 equations together give us $0 = \lambda (a + b + c)$. Since $a + b + c \ne 0$, this forces $\lambda = 0$. It is easy to see it is impossible for the first equation to work.

Next, let us switch to the problem of finding the minimum over $\bar{D}$. Since $\bar{D}$ is compact and $P : \bar{D} \to \mathbb{R} \cup \{ +\infty \}$ is continuous and bounded from below, $P$ reaches an absolute minimum somewhere in $\bar{D}$. This minimum cannot lie in $D$. At least one of $a,b,c$ need to vanish. However, we can't have more than one to vanish or $P$ will blow up. This means for the minimum, one and only one of $a,b,c$ is $0$. WOLOG, we will assume $0 = a < b < c$.

Let $t = \frac{b}{c} \in (0,1)$ and $u = t+t^{-1} \in (2,\infty)$, the target function $P$ becomes

$$\begin{align}P(u) = \frac{1}{b^2} + \frac{1}{(b-c)^2} + \frac{1}{c^2} &= \frac13\frac{b^2+c^2}{c^2}\left(\frac{1}{t^2} + \frac{1}{(1-t)^2} + 1\right)\\ &= \frac13(t + t^{-1})\left(t^{-1} + \frac{1}{t+t^{-1}-2} + t\right)\\ &= \frac{u}{3}\left(u + \frac{1}{u-2}\right) \end{align} \tag{*1} $$ Let $u_{min}$ be the $u$ corresponds to the absolute minimum. When $u = u_{min}$, we have $$P'(u) = \frac23\frac{(u-1)(u^2-3u+1)}{(u-2)^2} = 0 \quad\iff\quad u = 1, \frac{3\pm \sqrt{5}}{2}$$ Since $u_{min} > 2$, the last root is the only root relevant to us.

This means $u_{min} = \frac{3 + \sqrt{5}}{2}$ and for any admissible $u$, we have the inequality:

$$P = P(u) \ge P(u_{min}) = \frac{11+5\sqrt{5}}{6} \approx 3.696723314583158$$

The minimum value of $P$ over $\bar{D}$ (and the infimum over $D$) is $\frac{11+5\sqrt{5}}{6}$.

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