Remember the aim in archery?

Well, the inequality Hari prasad Vardarajan has given is correct and the following is the proof -

$PA = a, PB = b, PC = c, APB = x, BPC = y, CPA = z$

Using cosine rule,

${AB}^{2} + {BC}^{2} + {CA}^{2} = 2({a}^{2} + {b}^{2} + {c}^{2}) - 2abcos(x) - 2bccos(y) - 2cacos(z)$

$= 2({a}^{2} + {b}^{2} + {c}^{2}) - 2abcos(x) - 2bccos(y) - 2ca(cos(2\pi-(x+y)))$ [ $x+ y + z = 2\pi$ ]

$= 2({a}^{2} + {b}^{2} + {c}^{2}) - 2abcos(x) - 2bccos(y) - 2cacos(x)cos(y) + 2casin(x)sin(y)$

$= 2({a}^{2} + {b}^{2} + {c}^{2}) - cos(x)(2ab +2cacos(y))+ sin(x)(2casin(y)) - 2bccos(y)$

$= 2({a}^{2} + {b}^{2} + {c}^{2}) + cos(x)(-2ab -2cacos(y))+ sin(x)(2casin(y)) - 2bccos(y)$

We know $acos(x) + bsin(x)$ has maximum at $\sqrt{{a}^{2} + {b}^{2}}$

$\le 2({a}^{2} + {b}^{2} + {c}^{2}) + \sqrt{{(2ab + 2cacos(y))}^{2} + {(2casin(y))}^{2}} - 2bccos(y)$

$\le 2({a}^{2} + {b}^{2} + {c}^{2}) + \sqrt{4{a}^{2}{b}^{2} + 4{c}^{2}{a}^{2}{cos}^{2}(y) + 4{a}^{2}bccos(y) + 4{a}^{2}{c}^{2}{sin}^{2}(y)} - 2bccos(y)$

$\le 2({a}^{2} + {b}^{2} + {c}^{2}) + \sqrt{4{a}^{2}{b}^{2} + 4{c}^{2}{a}^{2}({cos}^{2}(y) + {sin}^{2}(y))+ 4{a}^{2}bccos(y)} - 2bccos(y)$

$\le 2({a}^{2} + {b}^{2} + {c}^{2}) + \sqrt{4{a}^{2}({b}^{2}+{c}^{2})+ 4{a}^{2}bccos(y)} - 2bccos(y)$

$\le 2({a}^{2} + {b}^{2} + {c}^{2}) + 2a\sqrt{{b}^{2}+{c}^{2}+ bccos(y)} - 2bccos(y)$

If we take derivative of the second part of the above equation involving $y$ w.r.t. $y$ , we get where the equality will hold and the maximum which will just be ${a}^{2} + {b}^{2} + {c}^{2}$

After adding, we get

${AB}^{2} + {BC}^{2} + {CA}^{2} \le 3({a}^{2} + {b}^{2} + {c}^{2})$

In a triangle ABC and for a point in space,there is inequality

$AB^2+BC^2+CA^2$ is less than or equal to 3( $PA^2+PB^2+PC^2$ )

Now if we take the point P to be the center then the rest is easy......