help

Diganta B., please don't forget to mention $\geq$ instead of $>$ in your question.

For the first one: Assume, by symmetry, $a \geq b \geq c$. Then, by Rearrangement inequality on the three sets $a^{2},b^{2},c^{2}$ , $a^{3},b^{3},c^{3}$ and $a^{3},b^{3},c^{3}$, we get

$a^{2}a^{3}a^{3}+b^{2}b^{3}b^{3}+c^{2}c^{3}c^{3}\geq a^{2}b^{3}c^{3}+a^{3}b^{2}c^{3}+a^{3}b^{3}c^{2}$

which gives $a^{8}+b^{8}+c^{8} \geq a^{3}b^{3}c^{3} (\frac {1}{a} + \frac {1}{b} + \frac {1}{c})$ And finally

$\frac {a^{8}+b^{8}+c^{8}}{a^{3}b^{3}c^{3}} \geq (\frac {1}{a} + \frac {1}{b} + \frac {1}{c})$

For the second question:

$\frac {a^{8}+b^{8}+c^{8}}{3} > (\frac {a+b+c}{3})^{8}$ is the same as

$(\frac {a^{8}+b^{8}+c^{8}}{3})^{\frac {1}{8}} > \frac {a+b+c}{3}$ and this nothing but the generalized mean with $M_{8}(a,b,c) > M_{1}(a,b,c)$. I hope you know about the generalized mean inequality.

Equality in both cases holds iff $a=b=c$

could you type these up? no offense, but i can't read any of the exponents

use AM-GM inequality

I think the second one is Titu's Lemma...

i. By AM-GM,

$\frac28a^8+\frac38b^8+\frac38c^8\geq a^2b^3c^3\\ \frac28b^8+\frac38c^8+\frac38a^8\geq b^2c^3a^3\\ \frac28c^8+\frac38a^8+\frac38b^8\geq c^2a^3b^3$

Adding these three inequalities and dividing by $a^3b^3c^3$ gives the desired inequality.

ii. This is a special case of the power-mean inequality on three variables.

Please type this up. I can't read it clearly.

1st one is obvious by weighted AM-GM. Second one is obvious by Jensen's inequality.

explain clearly

I thought we weren't supposed to put boring homework on this forum...