parmenides51 wrote: If $P$ is the area of a triangle $ABC$ with sides $a,b,c$, prove that $\frac{ab+bc+ca}{4P} \ge \sqrt3$ $$\frac{ab+bc+ca}{4P}=2\left(\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}\right)\geq 6\sqrt[3]{\frac{1}{\sin A \sin B \sin C}}\geq 4\sqrt{3} $$or $$\frac{ab+bc+ca}{4P}=2\left(\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}\right)\geq \dfrac{18}{\sin A+\sin B+\sin C} \geq 4\sqrt{3} $$or $$ab+bc+ca\geq (abc)^{\frac 23}\geq 4\sqrt 3S$$

Or use substitution $u-a=m$, $u-b=n$, and $u-c=k$, after which it boils down proving \[ \sum (m+n)(m+k)\ge 4\sqrt{3mnk(m+n+k)}. \]Using $(mn+nk+km)^2 \ge 3mnk(m+n+k)$, we then conclude $\sqrt{3mnk(m+n+k)}\le mn+nk+km$. Hence, it suffices to show $\sum (m+n)(m+k)\ge 4(mn+nk+km)$, which is equivalent to $m^2+n^2+k^2\ge mn+nk+km$, which is clear.

Prove the inequality $$a^2 + 2b^2 + 2c^2 \geq 8\sqrt{2} S$$where $ a$, $ b$, $ c$ are the sidelengths of a triangle $ ABC$, and $ S$ is its area.

Weitzenböck inequality