Let $P$ be a point in triangle $ABC$.
Prove that : $AP+BP+CP>2\sqrt[2]{\sqrt[2]{3}S}$ Such that $S$ is the area of the triangle.
I can only prove that if $P$ is a point inside any triangle $\Delta ABC$ then we have the following inequality(which I have proved using triangle inequality) ; $$s\leq AP+BP+CP\leq 2s,$$ where $s$ is the semi perimeter of the triangle. How can I solve the given problem. Thank you!
In the standard notation we'll prove that $$R_a+R_b+R_c\geq\sqrt{2\sqrt3S} $$ Indeed, let $\measuredangle{PAC}\leq120^{\circ}.$
Also, let $R_{A}^{60^{\circ}}$ is a rotation around $A$ by the angle $60^{\circ},$$ R_{A}^{60^{\circ}}(P)=P'$ and $R_{A}^{60^{\circ}}(B)=B'$.
Thus, by the triangle inequality $$R_a+R_b+R_c=PA+PB+PC=PP'+P'B'+PC\geq B'C=$$ $$=\sqrt{c^2+b^2-2bc\cos\measuredangle{B'AC}}=\sqrt{b^2+c^2-2bc\cos\left(60^{\circ}+\alpha\right)}=$$ $$=\sqrt{b^2+c^2-2bc\left(\frac{1}{2}\cdot\frac{b^2+c^2-a^2}{2bc}-\frac{\sqrt3}{2}\cdot\frac{2S}{bc}\right)}.$$ Thus, it's enough to prove that $$b^2+c^2-2bc\left(\frac{1}{2}\cdot\frac{b^2+c^2-a^2}{2bc}-\frac{\sqrt3}{2}\cdot\frac{2S}{bc}\right)\geq4\sqrt3S$$ or $$a^2+b^2+c^2\geq4\sqrt3S$$ or $$a^2+b^2+c^2\geq\sqrt{3\sum_{cyc}(2a^2b^2-a^4)}$$ or $$\sum_{cyc}(a^2-b^2)^2\geq0.$$ If $\measuredangle{PAC}>120^{\circ}.$
So $R_a+R_b+R_c>b+c$ and it remans to prove that: $$(b+c)^2>4\sqrt{3}S.$$ Can you end it now?