Geometric inequality involving the inradius 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:
$$p^2\ge 2\sqrt3 T+\frac {abc}{p}+r^2.$$
 A: 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.
