Geometric Application of Cauchy-Schwarz Inequality Problem

I have been struggling with this problem, and would like to prove the inequality using the Cauchy-Schwarz Inequality:

The vertices of a fixed triangle are $A$,$B$ and $C$, and $P$,$Q$ and $R$ lie on the line segments $BC$, $CA$ and $AB$ respectively. if $[XYZ]$ denotes the area of the triangle $XYZ$, prove that

Any help would be much appreciated!

Thanks

Let $m = CQ$, $n = QA$, $p = AR$, $q = RB$, $x = BP$, and $y = PC$. Also let $a = BC$, $b = AC$, and $c = AB$. Then:
$\sqrt{\dfrac{S_{AQR}}{S_{ABC}}} = \sqrt{\dfrac{np}{bc}} \leq \dfrac{1}{2}\cdot \left(\dfrac{n}{b} + \dfrac{p}{c}\right)$.
Similarly, we can obtain inequalities for the other two square-roots. Thus the sum $T$ of the left side satisfies:
$T \leq \dfrac{1}{2}\cdot \left(\dfrac{n}{b} + \dfrac{p}{c}\right) + \dfrac{1}{2}\cdot \left(\dfrac{q}{c} + \dfrac{x}{a}\right) + \dfrac{1}{2}\cdot \left(\dfrac{m}{b} + \dfrac{y}{a}\right) = \dfrac{3}{2}$. Done.