A hard geometric inequality Consider any point P in the interior of the triangle, with the triangle's vertices denoted $A, B, C$ and with the lengths of line segments denoted $PA, PB, PC$. If the perpendiculars from interior point $P$ to the sides of the triangle, intersecting the sides at $D, E, F$, we have for positive numbers $k_1, k_2, k_3$ and $t>0$:
$k_1 PA^t+k_2 PB^t+k_3 PC^t \ge 2 \sqrt{k_1 k_2 k_3} ( \frac{ PD^t }{ \sqrt{ k_1 } } + \frac{ PE^t }{\sqrt{ k_2 } } + \frac{ PF^t }{\sqrt{ k_3 } }) $
I have to prove that
$2x PA^2+2y PB^2+2z PC^2 > \frac {yz}{y+z} a^2+ \frac {zx}{z+x} b^2+ \frac {xy}{x+y} c^2$
for $x,y,z>0.$ We may consider $t=2$ and $ k_1=2x, k_2=2y, k_3=2z,$ but it seems hard to continue. Any help? Thank you
PS. By: Klamkin, M. S. (1975). Geometric inequalities via the polar moment of inertia. Mathematics Magazine, 48 (1): 44-46.
If $P$ is an interior point of a triangle $ABC$, and $a=BC, b=CA, c=AB,$ then
$(x+y+z)(x PA^2+y PB^2+z PC^2 ) \ge yz a^2+zx b^2+xy c^2.$
 A:  
The strategy here is to determine the configuration which maximizes the right side of the inequality compared to the left so you only have to deal with one case. My following answer is a rough cut at this and is by no means airtight. However, it is a strategy that can be made to work albeit plugging the holes and tidying up the whole thing.
The sum of sides $a,b,c$ are always longer than the sum of $PA,PB,PC$ so the question is, are they proportionally larger or smaller than the $2x,2y,2z$ left side factors compared to $\frac{yz}{(y+z)}$ etc right side factors?
Prove:        $2xPA^2 + 2yPB^2 + 2zPC^2 > \frac{yz}{(y+z)}a^2 + \frac{xz}{(x+z)}b^2 + \frac{xy}{(x+y)}c^2$
Consider Case 3, an equilateral triangle with point P at the centroid.
Let’s make $x, y, z = q$.  This will make $PA,PB,PC = 4q$ and $a,b,c = 4q√3$
$$2q(4q)^2*3 > \frac{q^2}{2q}(4q√3)^2*3$$
$$96q^3 > 72q^3$$
If this is the optimal case then this is proof enough. The following are a couple of cases to determine if this is indeed the optimal case.
For Case 1, making $x,y,z$ very small, which makes $2x,2y,2z$ small on the left but $\frac{yz}{(y+z)}$ etc. even smaller on the right. Hence it makes the right side smaller than the left compared to Case 3.
For Case 2, making {x,y} as large as possible but {z} very small, makes only one term of the left very small and the other two terms slightly larger. On the right it makes two terms very small and only one slightly larger. Again it makes the right side smaller than the left compared to Case 3.
Case 1 and Case 2 (there may be more required) do not increase the right side of the inequality compared to the left more so than Case 3 which so far is the optimal configuration in maximizing the ratio of right over left.
In the most optimal case, right/left $< 1$ hence: $$2xPA^2 + 2yPB^2 + 2zPC^2 > \frac{yz}{(y+z)}a^2 + \frac{xz}{(x+z)}b^2 + \frac{xy}{(x+y)}c^2$$
