Inequality involving the sum of square roots of distances from a fixed point in triangle to triangle sides Let $P$ be a point of a triangle $ABC$, and $d_1=dist(P,AB)$, $d_2=dist(P,BC)$, $d_3=dist(P,CA)$.
Prove that $\sqrt{d_1}+\sqrt{d_2}+\sqrt{d_3}\le \sqrt{\frac{9}{2}R}$

I re-define $d_1$ as $d(P,BC)=d(P,a)$ and so on for the sake of symmetry.
By the Cauchy-Schwarz inequality
$$\begin{eqnarray*}\sqrt{d_1}+\sqrt{d_2}+\sqrt{d_3} &=& \frac{1}{\sqrt{a}}\sqrt{ad_1}+\frac{1}{\sqrt{b}}\sqrt{bd_2}+\frac{1}{\sqrt{c}}\sqrt{cd_3}\\ &\leq& \sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(ad_1+bd_2+cd_3)}\\&=&\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)2\Delta}\\&=&\sqrt{\left(\frac{1}{2R\sin A}+\frac{1}{2R\sin B}+\frac{1}{2R\sin C}\right)R^2(\sin(2A)+\sin(2B)+\sin(2C))}\\&\leq&\sqrt{3R\left(\cos A+\cos B+\cos C\right)} \end{eqnarray*}$$
where $R(\cos A+\cos B+\cos C)=R+r$ by Carnot's theorem and $R\geq 2r$ by Euler's theorem.
