Maximum area of a right-angled triangle with a fixed perimeter - can it be done without calculus?

Question

The area, $A$, of a right-angled triangle with fixed perimeter $P$ will be maximised when the triangle is isosceles (i.e. is a multiple of the $1, 1, \sqrt{2}$ right-angled triangle). I know how to prove this using calculus, but I am interested to know whether there is an alternative proof that doesn't need advanced techniques. In particular, Simple proof that equilateral triangles have maximum area shows that a proof without calculus is possible in the general case (to show that the area of any triangle with fixed perimeter will be maximum when the triangle is equilateral), so I think a simple proof should also be possible in the right-angled case.

Answer 1

Let a , b , c be the sides of the right angle triangle right angle at C. $$ \therefore a^2+b^2=c^2 \,,where\,a+b+c=p\, ,\,perimeter\, of \,triangle$$ Using $AM\ge GM$ between $a^2,b^2$ we get $$\frac{a^2+b^2}{2} \ge (ab)$$ $$\therefore ab\le \frac{c^2}{2}$$ Hence maximum area is $A=\frac{1}{2}ab=\frac{c^2}{4}$ $$ $$ AM = GM holds when $a=b$ ttherfore $c=\sqrt{2}a$ Hence sides are $a,a,\sqrt{2}a$ hence $a=(\frac{p}{2}(2-\sqrt{2}))$ where p is constant perimeter.