Question

Assertion :In a triangle $ABC,\frac{a^2+b^2+c^2}{△}\ge 4\sqrt{3}$ Reason: If $a_i>0,i=1,2,3,...,n$ which are not identical, then
$\frac{a_1^m+a_2^m+...+a_n^m}{n}>{\left(\frac{a_1+a_2+...+a_n}{n}\right)}^m,$ if $m<0$ or $m>1$

• A. Both Assertion and Reason are individually true and Reason is the correct explanation of Assertion
• B. Both Assertion and Reason are individually correct but Reason is not the correct explanation of Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are individually true and Reason is the correct explanation of Assertion

We know that in any triangle
$\frac{s^2}{3\sqrt{3}}\ge △$ .............$\left(1\right)$
Now, $\frac{a^2+b^2+c^2}{3}\ge {\left(\frac{a+b+c}{3}\right)}^2$
$={\left(\frac{2s}{3}\right)}^2$
$=\frac{4}{9}{s}^2\ge \frac{4}{9}\times 3\sqrt{3}△$ from eqn$\left(1\right)$
$\therefore \frac{a^2+b^2+c^2}{△}\ge 4\sqrt{3}$