Maximization of function in 3 variables If $( x,y,z)$ be the lengths of perpendiculars from any interior point P of a triangle $ABC$ on sides $BC,CA$ and $AB$ respectively then find the minimum value of :
$$
x^2+ y^2 + z^2
$$
The sides of triangle being $a,b,c$.
I thought of using Lagrange's method of multipliers but am not able to find another function in terms of $x,y,z$ and $a,b,c$
Any help will be appreciated. Thanks. 
 A: Here is a solution without the method of Lagrange multipliers. Note that $ax+by+cz=2S$ where $S$ is the area of the triangle. Therefore, by Cauchy-Schwarz inequality we have 
$$\frac{4S^2}{a^2+b^2+c^2}= \frac{(ax+by+cz)^2}{a^2+b^2+c^2}\leq x^2+y^2+z^2.$$ The equality holds when $x:y:z=a:b:c$. The corresponding point is the radical center of the three Apollonian circles associated with the triangle.
A: As in $ x^2 + y^2 + z^2 $ all are squares , sum of squares is minimum when all terms (for which squares are taken) are equal .( refer How to prove the sum of squares is minimum?)  . So the point should be the orthocentre of triangle with equal altitudes . if altitude be 'h'. distance from orthocentre to base is $h/2$ , then minimum is $3(h^2)/4$ . height can be expresses in terms of area which is in terms of sides . so a relation is obtained as you wanted .
A: like Tunococ said,you can setup a triangle by your self first:
let $A(0,u),B(v,0),C(w,0), u>0,v>0,w<v$ then $a= v-w,b=\sqrt{u^2+w^2},c=\sqrt{u^2+v^2}$  
let point $P(p,q)$, then $x^2=q^2, y^2=\dfrac{(up+wq-uw)^2}{u^2+w^2},z^2=\dfrac{(up+vq-uv)^2}{u^2+v^2}$,since P is in the $\triangle ABC$,so $q\ge 0, up+wq-uw \le 0,up+vq-uv \le 0$
$u,v,w$ can be written in $f(a,b,c)$ so the question be come to find minimum of  $Q(p,q)=x^2+y^2+z^2$ ,now you can go further.  
Edit: it has a direct way as SB pointed out:
$ax+by+cz=2S$ so it is more easy to do.
