About a complex number $z\in S^1$ fulfilling an inequality Let $a,b,c$ be complex numbers, such that $\|a\|=\|b\|=\|c\|=1$. I have to show that there exists a complex number $z$, with $\|z\|=1$, such that:
$$\dfrac{1}{\|z-a\|^2}+\dfrac{1}{\|z-b\|^2}+\dfrac{1}{\|z-c\|^2}\le\dfrac{9}{4}$$
 A: A collection of hints.
We may study the behaviour of the function:
$$\begin{eqnarray*} f(\theta) &=& \frac{1}{\|e^{i\theta}-e^{i\alpha}\|^2}+\frac{1}{\|e^{i\theta}-e^{i\beta}\|^2}+\frac{1}{\|e^{i\theta}-e^{i\gamma}\|^2}\\&=&\frac{1}{4\sin^2\frac{\alpha-\theta}{2}}+\frac{1}{4\sin^2\frac{\beta-\theta}{2}}+\frac{1}{4\sin^2\frac{\gamma-\theta}{2}}\end{eqnarray*}$$
over the interval $[0,2\pi]$. We may notice that for any $\varphi\in[0,2\pi]$ the measure of the subset of $[0,2\pi]$ for which
$$ \frac{1}{4\sin^2\frac{\varphi-\theta}{2}}\geq 1 $$
is exactly $\frac{2\pi}{3}$, hence for any $\varepsilon>0$ the measure of the subset of $[0,2\pi]$ for which
$$ f(\theta)\leq 3(1+\varepsilon)$$
is positive, so:
$$ \forall \varepsilon>0,\;\exists\, \theta\in[0,2\pi]:\quad f(\theta)\leq 3(1+\varepsilon).$$
Now you just need to improve a bit this argument.
We may try this approach: assume that $ABC$ is a triangle inscribed in a unit circle, with the $A$ vertex being associated with the greatest angle. If we take the symmetric of $A$ with respect to the centre of the previous circle, or the midpoint of the $BC$-arc not containing $A$ $\phantom{}^{(*)}$, we get a reasonable $z$. Another good idea may be to exploit Ptolemy's theorem or a circular inversion centered at $z$ to send $A,B,C$ into three collinear points and deal with the sum of three squared distances:


(*): Just to restate it clearly, the inequality is trivial for obtuse triangles (the constant $\frac{9}{4}$ can be replaced by $\frac{3}{2}$). For acute triangles, if we take $z$ as the midpoint of the minor arc of the longest chord, $z$ fulfils the initial inequality, and such inequality is clearly optimal since equality is attained by the equilateral triangle. 

A: Let $f(z)=\frac{1}{|z-a|^2}+\frac{1}{|z-b|^2}+\frac{1}{|z-c|^2}$. The points $a,b,c$ divide the unit circle $|w|=1$ into three parts. We claim that we can take $z$ to be $d$ to be the midpoint of the longest part. Wlog that is $a$ to $b$. 

If $a,b,c$ all lie on a semicircle, then $|c-d|>|a-d|=|b-d|\ge\sqrt2$, so $f(d)<\frac{3}{2}<\frac{9}{4}$. So we can assume the difference between the arguments of $a,b$ is less than $\pi$.
The distance between two points on the unit circle whose arguments differ by $\theta$ is $2\sin\frac{\theta}{2}$. For the points $a,b$ we have $\theta=\frac{2\pi}{3}+2x$ for some $0<x<\frac{\pi}{6}$. So the argument difference between $a,d$ is $\frac{\pi}{3}+x$. Hence $|a-d|=|b-d|=2\sin(\frac{\pi}{6}+\frac{x}{2})$. 
Since the argument difference between $a,c$ and between $b,c$ is less than $\theta$, the argument of $c$ differs by at most $3x$ from the antipodal point to $d$, so the argument difference between $d,c$ is at least $\pi-x$. Hence $|c-d|\ge2\sin(\frac{\pi}{2}-\frac{3x}{2})$.
So we are home provided $$\frac{2}{\sin^2(\frac{\pi}{6}+\frac{x}{2})}+\frac{1}{\sin^2(\frac{\pi}{2}-\frac{3x}{2})}\le9$$ for all $0<x<\frac{\pi}{6}$. The first term decreases from 8 at $x=0$ to 4 at $x=\frac{\pi}{6}$. The second term increases from 1 at $x=0$ to 2 at $x=\frac{\pi}{6}$.
Differentiating we get $-2\cot(\frac{\pi}{6}+\frac{x}{2})\csc^2(\frac{\pi}{6}+\frac{x}{2})+3\sec^2\frac{3x}{2}\tan\frac{3x}{2}$. Both $\cot$ and $\csc$ are decreasing over the range, so the first term is increasing. Both $\sec$ and $\tan$ are increasing over the range. So the derivative is increasing. At $x=0$ it is $-8\sqrt3$. So it has at most one stationary point in the range, which would be a minimum. Hence its maximum value must be at one of the two endpoints. Those are 9,6. So the maximum is 9 (corresponding to $a,b,c$ forming an equilateral triangle), as required.
