Properties Of Triangles Ques 39
If $\Delta$ is the area of a triangle with side lengths $a, b, c$, then show that
$ \Delta \leq \frac{1}{4} \sqrt{(a+b+c) a b c} $
Also, show that the equality occurs in the above inequality if and only if $a=b=c$.
$(2001,6 M)$
$
\Delta \leq \frac{1}{4} \sqrt{(a+b+c) a b c}
$ $
\begin{aligned}
& \Rightarrow \quad \frac{1}{4 \Delta} \sqrt{(a+b+c) a b c} \geq 1 \\
& \Rightarrow \quad \frac{(a+b+c) a b c}{16 \Delta^{2}} \geq 1 \\
& \Rightarrow \quad \frac{2 s a b c}{16 \Delta^{2}} \geq 1 \\
& \Rightarrow \quad \frac{s a b c}{8 \cdot s(s-a)(s-b)(s-c)} \geq 1 \\
& \Rightarrow \quad \frac{a b c}{8(s-a)(s-b)(s-c)} \geq 1 \\
& \Rightarrow \quad \frac{a b c}{8} \geq(s-a)(s-b)(s-c)
\end{aligned}
$ Now, put $s-a=x \geq 0, s-b=y \geq 0, s-c=z \geq 0$ $
\begin{aligned}
s-a+s-b & =x+y \\
2 s-a-b & =x+y \\
c & =x+y
\end{aligned}
$ Similarly, $a=y+z, b=x+z$ $\Rightarrow \quad \frac{(x+y)}{2} \cdot \frac{(y+z)}{2} \cdot \frac{(x+z)}{2} \geq x y z$ which it true. Now, equality will hold, if $x=y=z$ $\Rightarrow a=b=c$ $\Rightarrow$ Triangle is equilateral.Show Answer
Solution:
BETA
