In equilateral ABC , Proving : $3(a^4 + b^4 + c^4 + s^4 ) = (a^2 + b^2 + c^2 + s^2 )^2$ My question is that:
In equilateral $\triangle ABC$,  $AB= s$ 
There is a point P in same plane , the distance to three vertices are $a, b, c $
then
$$ 3(a^4 + b^4 + c^4 + s^4 ) = (a^2 + b^2 + c^2 + s^2 )^2 $$
Is this equation true? If it is true How can I prove?

This is not a solution, although the equality seems to be true. One way to approach the problem might be to work with the points themselves as opposed to distances.
We can put your triangle on a coordinate plane and say that Point A is at $(-\frac{s}{2} , 0)$, Point B is at $(\frac{s}{2}, 0)$, and Point C is at $(0, \frac{s\sqrt{3}}{2})$.
Then, if we say Point P is at $(x, y)$, we have that
$$a = \sqrt{\Big(x+\frac{s}{2}\Big)^2 +y^2}$$
$$b = \sqrt{\Big(x-\frac{s}{2}\Big)^2 +y^2}$$
$$c = \sqrt{x^2 +\Big(y-\frac{s\sqrt{3}}{2}\Big)^2}$$
Now, foiling out the RHS of what you want to prove, we get that:
$$(a^2 + b^2 + c^2 + s^2)^2 = (a^4 +b^4 + c^4 + s^4) + 2a^2b^2 + 2a^2c^2 + 2a^2s^2 + 2b^2c^2 + 2c^2s^2 + 2b^2s^2$$
So, in other words, you question is equivalent to showing that: 
$$a^4 + b^4 + c^4 + s^4 = a^2b^2 +a^2c^2 +a^2s^2 + b^2c^2 +c^2s^2 + b^2s^2  $$
Using the definitions of $a, b$ and $c$ from above might be a method of showing the right relationships to prove this is true. For example, $a$ and $b$ clearly have a close relationship as $a^2 - b^2 = 2sx$.
