Explanation

(a)
Now, $3\left( a^{2} + b^{2} + c^{2} \right) - (a + b + c)^{2}$
$= 2(a^{2} + b^{2} + c^{2} - bc - ca - ab)$
$= (b - c)^{2} + (c - a)^{2} + (a - b)^{2} \geq 0$
$\Rightarrow 3\left( a^{2}{+ b}^{2} + c^{2} \right) \geq (a + b + c)^{2} > 9$
$\Rightarrow a^{2} + b^{2} + c^{2} > 3 \Rightarrow \$(a)holds
Now, $a^{6} + b^{6} \geq 12a^{2}b^{2} - 64$
If $a^{6} + b^{6} + 64 \geq 12a^{2}b^{2}$
$ie,\ \ \ a^{6} + b^{6} + 2^{6} \geq {3 \bullet 2}^{2} \bullet a^{2}b^{2}$
$ie,if\ \ \frac{a^{6} + b^{6} + 2^{6}}{3} \geq \left( 2^{6}a^{6}b^{6} \right)^{1/3}\ \ \ (\because AM \geq GM)$
$\Rightarrow (b)$does not hold
Again, since $AM \geq HM$
$\therefore\frac{a + b + c}{3} \geq \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$
$\Rightarrow \frac{\alpha}{3} \geq \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$
$\Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq \frac{9}{\alpha}$
$\Rightarrow (c$ ) does not hold