Equilateral triangle and another triangle with same perimeter. Which has larger area? There is an equilateral triangle with sides $a$ and another triangle with sides $p,q,r$, both having the same perimeter $S$.
How can we mathematically show which of them has a larger area?

Have you heard of Heron's formula? Especially on the form
$$
A = \sqrt{\frac S2\left(\frac S2-a\right)\left(\frac S2-b\right)\left(\frac S2-c\right)},\quad a, b, c\text{ are the sides of the triangle}
$$we come very close to a full solution. What we need to get all the way is the AM-GM inequality, which states that for any three positive numbers $k,l,m$, we have
$$
\frac{k+l+m}{3}\geq \sqrt[3]{klm}
$$

Since you asked, here is a full work-out:
For the equilateral triangle, we have $k = l = m = (S/2 - a)$, and the AM-GM inequality is actually an equality:
$$
\frac{S/2 - a + S/2 - a + S/2 - a}{3} = \sqrt[3]{(S/2 - a)^3}\\
\frac{3S/2 - 3a}{3} = \sqrt[3]{(S/2 - a)^3}\\
\frac{3S/2 - S}{3} = \sqrt[3]{(S/2 - a)^3}\\
\frac S6 = \sqrt[3]{(S/2-a)^3}\\
\left(\frac S6\right)^3 = \left(\frac{S}2 - a\right)^3
$$
(Although, that could've been worked out without using AM-GM.)
Now, for a non-equilateral triangle, we set $k = (S/2 - p), l = (S/2 - q)$ and $m = (S/2 - r)$, and the AM-GM inequality is a strict inequality:
$$
\frac{S/2 - p + S/2 - q + S/2 - r}{3} > \sqrt[3]{(S/2 - p)(S/2 - q)(S/2 - r)}\\
\frac{3S/2 - (p+q+r)}{3} > \sqrt[3]{(S/2 - p)(S/2 - q)(S/2 - r)}\\
\frac{3S/2 - S}{3} > \sqrt[3]{(S/2 - p)(S/2 - q)(S/2 - r)}\\
\frac S6 > \sqrt[3]{(S/2 - p)(S/2 - q)(S/2 - r)}\\
\left(\frac S6\right)^3 > \left(\frac S2 - p\right)\left(\frac S2 - q\right)\left(\frac S2 - r\right)
$$
Now insert these two different products into Heron's formula, and you can see that the equilateral triangle gives the bigger area.
