# The question is below?

## If $\Delta$ is the area of a triangle with side lengths $a , b , c$ then show that $\Delta \le \frac{1}{4} \sqrt{\left(a + b + c\right) \cdot a b c}$. Also show that equality occurs if and only if $a = b = c$.

Jun 6, 2018

Considering the Heron's formula of area of the triangle $A B C$
$\Delta = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$, where $a , b \mathmr{and} c$ are three sides of the triangle and its semi perimeter is $s$ i.e.$2 s = a + b + c$
Now let

$s - a = x , s - b = y \mathmr{and} s - c = z$

Now each of $x , y \mathmr{and} z$ is greater than zero as sum of two sides of a triangle is greater than third side.

So

$x + y = 2 s - a - b = a + b + c - a - b = c$

Similarly

$y + z = 2 s - b - c = a$

And

$z + x = 2 s - c - a = b$

Now we know $A M > G M$

So $\frac{x + y}{2} \ge \sqrt{x y}$

$\frac{y + z}{2} \ge \sqrt{y z}$

$\frac{z + x}{2} \ge \sqrt{z x}$

So multiplying we get

$\sqrt{x y \cdot y z \cdot z x} \le \frac{1}{8} \left(x + y\right) \left(y + z\right) \left(z + x\right)$

$\implies \sqrt{{x}^{2} {y}^{2} {z}^{2}} \le \frac{1}{8} \left(x + y\right) \left(y + z\right) \left(z + x\right)$

$\implies x y z \le \frac{1}{8} \left(x + y\right) \left(y + z\right) \left(z + x\right)$

$\implies \left(s - a\right) \left(s - b\right) \left(s - c\right) \le \frac{1}{8} c b a$

$\implies s \left(s - a\right) \left(s - b\right) \left(s - c\right) \le \frac{1}{8} s c b a$
$\implies s \left(s - a\right) \left(s - b\right) \left(s - c\right) \le \frac{1}{8} \cdot \frac{1}{2} \left(a + b + c\right) a b c$

$\implies \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)} \le \sqrt{\frac{1}{16} \left(a + b + c\right) a b c}$

$\implies \Delta \le \frac{1}{4} \sqrt{\left(a + b + c\right) a b c}$

We know for an equilateral triangle of side $a$
$\Delta = \frac{\sqrt{3}}{4} {a}^{2}$. So equality of the above relation is satisfied when $a = b = c$