Geometric inequality involving the inradius

Question

For the $\triangle ABC$, let $T$ be the area of the triangle, $a,b,c$ its sides, $p$ the semiperimeter and $r$ the inradius. Prove the following inequality: $$p^2\ge 2\sqrt3 T+\frac {abc}{p}+r^2.$$

Answer 1

Using the known formula $pr=T$, we rewrite the inequality as follows $$ p^2 - \frac{abc}{p} - \frac{T^2}{p^2} \ge 2 \sqrt{3} T$$ Now, using Heron's formula, $$ \begin{align*} \mathrm{LHS} &= \frac{1}{4} (a+b+c)^2 - \frac{2abc}{a+b+c} - \frac{(a+b-c)(a-b+c)(-a+b+c)}{4(a+b+c)} \\ &=\frac{1}{4} (a+b+c)^2 - \frac{8abc+(a+b-c)(a-b+c)(-a+b+c)}{4(a+b+c)}\\ &=\frac{1}{4} (a+b+c)^2 - \frac{(a+b+c)\left(2ab+2bc+2ca-a^2-b^2-c^2\right)}{4(a+b+c)}\\ &=\frac{1}{2} \left(a^2+b^2+c^2 \right) \end{align*}$$ So, the inequality becomes $$ a^2+b^2+c^2 \ge 4 \sqrt{3} T $$ Let $\gamma \colon \!= \widehat{ACB}$. Then, by the Law of cosines $$ c^2 = a^2 + b^2 -2ab \cos \gamma $$ By the trigonometric formula of the area of a triangle $$ T= \frac{1}{2} ab \sin\gamma $$ The inequality becomes then $$ 2 \left( a^2+b^2 - ab \left(\cos\gamma + \sqrt{3} \sin\gamma \right) \right)\ge 0$$ And this is true since $$\begin{align*} a^2+b^2 - ab \left(\cos\gamma + \sqrt{3} \sin\gamma \right) &= a^2+b^2- 2 ab \sin\left(\gamma+ \frac{\pi}{6} \right) \\ & \ge a^2+b^2- 2 ab = (a-b)^2 \ge 0 \end{align*}$$ And you can also see that the equality holds iff $a=b$ and $\gamma= \frac \pi 3$, that is when the triangle is equilateral.