Answer: Option A
:
A, B, C, and D
(a), (b), (c), (d)
(a) $\frac{a^4+b^4+c^4}{3}>{\left(\frac{a+b+c}{3}\right)}^4$
or $\left(\frac{a+b+c}{3}\right){\left(\frac{a+b+c}{3}\right)}^3>\frac{a+b+c}{3}\left[\right(abc{)}^{\frac{1}{3}}{]}^3$;
$\because A.M.>G.M.$
or $>\left(\frac{a+b+c}{3}\right)$
$\therefore a^4+b^4+c^4>abc(a+b+c)$
(b) As above
(c) $\left(\frac{a^5+b^5+c^5}{3}\right)>{\left(\frac{a+b+c}{3}\right)}^5$
or $>{\left(\frac{a+b+c}{3}\right)}^3{\left(\frac{a+b+c}{3}\right)}^2$
or $>\left[\right(abc{)}^{\frac{1}{3}}{]}^3\left(\frac{a^2+b^2+c^2+2ab+2bc+2ca}{9}\right)$
But we know that $a^2+b^2+c^2>ab+bc+ca$ by result of two by two rule.
$\therefore \frac{a^5+b^5+c^5}{3}>abc\frac{3(ab+bc+ca)}{9}$ etc.
(d) $a^8+b^8+c^8>a^2{b}^2{c}^2(bc+ca+ab)$
Now $\frac{a^8+b^8+c^8}{3}>{\left(\frac{a+b+c}{3}\right)}^8$
or $>{\left(\frac{a+b+c}{3}\right)}^6{\left(\frac{a+b+c}{3}\right)}^2>\left[\right(abc{)}^{\frac{1}{3}}{]}^6\left[\frac{a^2+b^2+c^2+2ab+2bc+2ca}{9}\right]$
$\because A.M.>G.M$
But by two by two rule
$a^2+b^2+c^2>ab+bc+ca$
$\therefore \frac{a^8+b^8+c^8}{3}>a^2{b}^2{c}^2\frac{(3ab+3bc+3ca)}{9}$
$\therefore a^8+b^8+c^8>a^2{b}^2{c}^2(ab+bc+ca)$
or $\frac{a^8{b}^8{c}^8}{a^3{b}^3{c}^3}>\frac{ab+bc+ca}{abc}$ or $>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
(e) $\frac{b^2+c^2}{2}>{\left(\frac{b+c}{2}\right)}^2\therefore \frac{b^2+c^2}{b+c}>\frac{b+c}{2}$
Write similar inequalities and add.