Question

Let a, b, c be the sides of a triangle whose perimeter is $p$ and area is $A$, then

• A. $p^3\le 27(a+c-b)(b+c-a)(a+b-c)$
• B. $p^2\le 3(a^2+b^2+c^2)$
• C. $a^2+b^2+c^2\ge 4\sqrt{3}A$
• D. $p^4\le 256A^2$

Answer 1

Answer: (B), (C)

The correct options are
B $p^2\le 3(a^2+b^2+c^2)$
C $a^2+b^2+c^2\ge 4\sqrt{3}A$

In $\Delta ABC$, $b+c-a>0$, $c+a-b>0$, $a+b-c>0$
So, $\frac{(b+c-a)+(c+a-b)+(+a+b-c)}{3}\ge \left\{\right(b+c-a\left)\right(c+a-b\left)\right(a+b-c){\}}^{\frac{1}{3}}$
$\Rightarrow p^3\ge 27(b+c-a)(c+a-b)(a+b-c)$
Also $\frac{(a+b+c)+(b+c-a)+(c+a-b)+(a+b-c)}{4}\ge {\left[\begin{array}{c}(a+b+c)(b+c-a)\\ (c+a-b)(a+b-c)\end{array}\right]}^{\frac{1}{4}}$
$\frac{2p}{4}\ge \left(\right(16A{)}^2{)}^{\frac{1}{4}}\Rightarrow p^2\ge 64A$
and $A\le \frac{\sqrt{3}}{4}{\left(\frac{p}{3}\right)}^2$ (for equilateral triangle $a=b=c=\frac{p}{3}$)
$p^2=(a+b+c{)}^2=(a^2+b^2+c^2)+2(ab+bc+ca)$
$p^2=3(a^2+b^2+c^2)-2(a^2+b^2+c^2-ab-bc-ca)$
$p^2\le 3(a^2+b^2+c^2)$ (equality holds iff a=b=c)
$A\le \frac{\sqrt{3}}{4}\frac{(a^2+b^2+c^2)}{3}\Rightarrow a^2+b^2+c^2\ge 4\sqrt{3}A$