Is the Weitzenbock inequality equivalent to the Roland inequality? Is the Weitzenbock triangle inequality $$a^2 + b^2 + c^2 \geq 4\sqrt{3}S$$ equivalent to the Roland inequality $$ab + bc + ca \geq 4\sqrt{3}S?$$--prophetes.ai

Is the Weitzenbock inequality equivalent to the Roland inequality? Is the Weitzenbock triangle inequality $$a^2 + b^2 + c^2 \geq 4\sqrt{3}S$$ equivalent to the Roland inequality $$ab + bc + ca \geq 4\sqrt{3}S?$$

As stated in the comments, Roland inequality is sharper since $a^2+b^2+c^2\geq ab+ac+bc$ holds by the rearrangement inequality or simply from $(a-b)^2+(a-c)^2+(b-c)^2\geq 0$. About proofs, I have outlined some proofs of Weitzenbock inequality here (among them, a nice proof without words). The inequality

$$ ab+ac+bc \geq 4\Delta\sqrt{3} $$ can be proved by invoking Heron's formula, Ravi substitution and Muirhead's inequality, or simply by exploiting $ab=\frac{2\Delta}{\sin C}$ and the convexity of $\frac{1}{\sin\theta}$ over $(0,\pi)$.