By carnot theorem we have $ BL^{2}+CM^{2}+AN^{2} = BN^{2}+CL^{2}+AM^{2} =\frac{1}{2}(BL^{2}+CL^{2}+CM^{2}+AM^{2}+AN^{2}+BN^{2})$ But $BL^{2}+CL^{2}\geq \frac{1}{2}(BL+CL)^{2}=\frac{1}{2}BC^{2}$ $CM^{2}+AM^{2}\geq \frac{1}{2}CA^{2}$ $AN^{2}+BN^{2}\geq \frac{1}{2}AB^{2}$ Hence $BL^{2}+CM^{2}+AN^{2}\geq \frac{1}{2}(AB^{2}+BC^{2}+CA^{2})$ The equality holds when $P$ is circumcenter of $\Delta ABC$

we can write it as $PB^2- PL^2 +PC^2- PM^2+PQ^2-PN^2 = X $ by erdos–mordell inequality $PL^2 +PM^2+PN^2 \le \frac{1}{2}(PB^2+PC^2+PA^2)$ Since For $X$ to be minimum we want $PL^2 +PM^2+PN^2$ to be maximum hence the proof follows

This problem is nice and easy: Let $BL=x,CM=y,AN=z$.Then, $0=\sum{BP^2-CP^2}=\sum{x^2-(a-x)^2} \Rightarrow x^2+y^2+z^2=(a-x)^2+(b-y)^2+(c-z)^2 =\frac{1}{2}(x^2+(a-x)^2+y^2+(b-y)^2+z^2+(c-z)^2) \ge \frac{a^2+b^2+c^2}{4}$ where we have used the fact that $(a^2+b^2) \ge \frac{(a+b)^2}{2}$. Equality holds if and only if $x=(a-x),y=(a-y),z=(a-z)$ i.e if and only if $P$ is the circumcenter of $\triangle{ABC}$.

Notice that $2BL^2+2CL^2\ge (BL+CL)^2=BC^2$ with equality when $BL=CL.$ Similarly, $2CM^2+2AM^2\ge CA^2$ and $2AN^2+2BN^2\ge AB^2$ with equality when $CM=AM$ and $AN=BN,$ respectively. By Carnot's Theorem we have $BL^2+CM^2+AN^2=CL^2+AM^2+BN^2,$ so $$BL^2+CM^2+AN^2\ge \frac{BC^2+CA^2+AB^2}{4}$$with equality when $L,M,N$ are the midpoints of $BC,CA,AB,$ respectively, or when $P$ is the circumcenter of $\triangle ABC.$ $\square$

After application of Carnot's theorem on equality $$(AN+NB)^2+(BL+LC)^2+(CM+MA)^2=a^2+b^2+c^2,$$we would like to maximise $AN\cdot NB+BL\cdot LC+CM\cdot MA$, but note that if $x+y=c$, then $x\cdot y=(c-x)\cdot x\leq \frac{c^2}{4}$, where equality holds iff $x=y=\frac{c}{2}$. This all in all means that $M,N,L$ must be midpoints in order to minimise $BL^2+CM^2+AN^2$, i.e. $P\equiv O$, the circumcenter of $\triangle ABC$. $\blacksquare$

We first focus on one side: \[\frac{BC^2}{2} = \frac{BL^2 + 2 BL \cdot CL + CL^2}{2} \leq BL^2 + CL^2.\] We can obtain symmetric inequalities for $AB$ and $AC$ to get \[\frac 14 \left(AB^2 + BC^2 + CA^2\right) \leq 2 \left(BL^2 + CL^2 + CM^2 + AM^2 + AN^2 + BN^2\right) = BL^2 + CM^2 + AN^2.\] Note that equality holds when $BL = CL$, $CM = AM$, and $AN = BN$, or $L$, $M$, and $N$ are midpoints. Thus $P$ must be the $\boxed{\text{circumcenter}}$. [asy][asy] pair A, B, C, P, L, M, N; A = dir(120); B = dir(210); C = dir(330); P = (-.2, 0);L = foot(P, B, C); M = foot(P, C, A); N = foot(P, A, B); draw(A--B--C--cycle); draw(P--L^^P--M^^P--N); label("$A$", A, dir(90)); label("$B$", B, SW); label("$C$", C, SE); label("$P$", P, dir(90)); label("$L$", L, S); label("$M$", M, NE); label("$N$", N, NW); [/asy][/asy]