First , $\triangle ASR \sim \triangle ABC$ we have $\frac{x}{a}=\frac{h_a-x}{h_a}$ hence $x=\frac{ah_a}{a+h_a}$, $\sum \frac{1}{x}=\sum \frac{1}{h_a}+\sum \frac{1}{a}$ Hence the inequality is equive to :$\sum a +\sum \frac{2F}{a} <=(2+\sqrt{3})\frac{a+b+c}{2}$ $\Leftrightarrow$ $\sum \frac{2F}{a} <=\sqrt{3}\frac{a+b+c}{2}$ $\Leftrightarrow$ $\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})<=\sqrt{3}(a+b+c)$ Assume : $a+b-c=z , a-b+c=y , -a+b+c=x$ The inequality can be written as $2\sqrt {xyz}(\sum \frac{1}{x+y})<=\sqrt{3(x+y+z)}$ $\Leftrightarrow$ $4(\sum \frac{1}{x+y})^2<=3(\sum \frac{1}{xy})$ $LEFT<=(\sum \frac{1}{\sqrt{xy}})^2<=(1+1+1)(\sum \frac{1}{xy})=RIGHT$ (AM-GM and Cauchy inequality) QED