Question

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

• A. $P^3\ge 27(b+c-a)(c+a-b)(a+b-c)$
• B. $P^2\le 3(a^2+b^2+c^2)$
• C. $a^2+b^2+c^2\ge 4\sqrt{3}A$
• D. none of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A $P^3\ge 27(b+c-a)(c+a-b)(a+b-c)$
B $P^2\le 3(a^2+b^2+c^2)$
C $a^2+b^2+c^2\ge 4\sqrt{3}A$

In $△ABC,b+c-a>0,c+a-b>0,a+b-c>0$
So, $\frac{(b+c-a)+(c+a-b)+(a+b-c)}{3}\ge \{(b+c-a)(c+a-b)(a+b-c){\}}^{\frac{1}{3}}$
$\Rightarrow \frac{a+b+c}{3}\ge \{(b+c-a)(c+a-b)(a+b-c){\}}^{\frac{1}{3}}$
Cubing both sides we get
$\Rightarrow P^3\ge 27(b+c-a)(c+a-b)(a+b-c)$
Also,$\frac{(a+b+c)+(b+c-a)+(c+a-b)+(a+b-c)}{4}>{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}^{\frac{1}{4}}$
$\Rightarrow \frac{2a+2b+2c}{4}\ge {(a+b+c)(b+c-a)(c+a-b)(a+b-c)}^{\frac{1}{4}}$
$\Rightarrow \frac{2P}{4}>{16A}^{\frac{1}{4}}$
$\Rightarrow \frac{P}{2}>2A^{\frac{1}{4}}$
$\Rightarrow P>4A^{\frac{1}{4}}$
$\Rightarrow P^4=256A$
For a given parameter, equilateral triangle has the largest area, so the area of triangle
$A\ge \frac{\sqrt{3}}{4}{\left(\frac{P}{3}\right)}^2$
for an equilateral triangle $a=b=c=\frac{P}{3}$
Now, $P^2={(a+b+c)}^2$
$=a^2+b^2+c^2+2(ab+bc+ca)$
$=3(a^2+b^2+c^2)-2(a^2+b^2+c^2-ab-bc-ca)$
$\le 3(a^2+b^2+c^2)$
equality holds if and only if $a=b=c$
Thus, $A\le \frac{\sqrt{3}}{4}\frac{a^2+b^2+c^2}{3}$
$\Rightarrow a^2+b^2+c^2\ge 4\sqrt{3}A$