Question

If a triangle with sides (in units) of measure $a,b$ and $c$ has an area of ${}^{\prime }{A}^{\prime }squnits$ and a triangle with sides measuring $a+b,b+c,c+a$ has an area of ${}^{\prime }{1}^{\prime }squnits$, then prove that $T\ge 4A$.

Answer 1

$a,b,c\sqrt{\frac{a+b+c}{2}(-\frac{a+b+c}{2})\left(\frac{a-b+c}{2}\right)\left(\frac{a+b-c}{2}\right)}$

$a+b,b+c,c+aA=\sqrt{(a+b+c)(-\frac{a-b+b+c}{2})\left(\frac{a+b-c-a}{2}\right)}$

$1=\sqrt{(a+b+c)(b+c)(c+a)(a+b)}$