Finding the triangle with the maximum area with a given perimeter 
Question: Of all triangles with a given perimeter, find the triangle with the maximum area. Justify your answer. 

My approach: Let us have any $\Delta ABC$ such that the sides opposite to $A$ is of length $a$, the side opposite to $B$ is of length b and the side opposite to $C$ is of length $c$. Now since the perimeter to $\Delta ABC$ is fixed, thus we must have $P=a+b+c$ to be constant. 
Now fix any side of the triangle, let us fix $BC$. Therefore $b+c=P-a$, which implies that $b+c$ is constant. Thus the locus of the point $A$ must be an ellipse having one of it's axis as the side $BC$. Now let us select any point $A$ on the ellipse and drop a perpendicular to the axis $BC$. Let it meet the major axis at $P$. Now let $AP=h$. Therefore, the area of the $\Delta ABC=\frac{1}{2}.h.BC=\frac{1}{2}.h.a.$ Now since $\frac{1}{2}a$ is constant, implies the area of $\Delta ABC$ can be maximized by maximizing $h$. Now clearly $h$ attains it's maximum value when it coincides with the other axis of the ellipse under consideration, that is when $AB=AC$. Thus $\Delta ABC$ must be isosceles with $AB=AC$ to get a maximum value of the area of $\Delta ABC$. 
Thus it is clear that the triangle which will have the maximum area must be one of the isosceles triangles $ABC$ having $BC$ as the base. 
Thus in any such $\Delta ABC$, we must have $a+b+c=a+2b \hspace{0.2 cm}(\because b=c)=P\implies b=\frac{1}{2}(P-a).$ Thus by Heron's formula we have $$|\Delta ABC|=\sqrt{\frac{P}{2}\left(\frac{P}{2}-a\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)}=\frac{a\sqrt{P}}{4}\sqrt{P-2a}.$$
Now to obtain the condition for maximizing $|\Delta ABC|$ we have to check when $\frac{d}{da}\frac{a\sqrt{P}}{4}\sqrt{P-2a}=0.$ Now $$\frac{d}{da}\frac{a\sqrt{P}}{4}\sqrt{P-2a}=0\\\iff (P-2a)^{1/2}-a(P-2a)^{-1/2}=0\\\iff P-2a=a\\\iff 3a=P\\\iff a=\frac{P}{3}.$$ 
Now note that $\frac{d^2}{da^2}|\Delta ABC|<0$, which implies that $|\Delta ABC|$ attain it's maximum value when $a=\frac{P}{3}.$ This implies that $b=c=\frac{P}{3}$. Thus we have $a=b=c=\frac{P}{3}$. Therefore, $|\Delta ABC|$ is maximized if and only if $a=b=c$, i.e, the triangle is equilateral. 
Can someone check if this solution is correct or not? And other solutions are welcomed. Please ensure that the solutions are based on geometry and one-variable calculus. This problem can be solved using Lagrange multipliers or multi-variable calculus, but I do not want a solution using the same. 

Use the Heron's formula
$$A =\sqrt{s(s-a)(s-b)(s-c)}, \>\>\>\>\> s=\frac p2$$
and apply the AM-GM inequality to get
$$A =\sqrt{s(s-a)(s-b)(s-c)} \le \left[ s \left( \frac{3s-(a+b+c)}{3} \right) ^3\right]^{1/2}
= \frac{s^2}{3\sqrt3}=\frac{p^2}{12\sqrt3}$$
where the equality, or the maximum area, occurs at $a=b=c=\frac p3$.
