It may be slightly long ..but this is one method i could immediately think of ..

$\mbox{To prove} \ \sum \frac{1}{\sqrt{1+\frac{8bc}{a^2}}}\geq1 \\ \\ \mbox{Now} \ \sum \frac{1}{\sqrt{1+\frac{8bc}{a^2}}}\geq \frac{9}{\sum \sqrt{1+\frac{8bc}{a^2}}} \ \mbox{Using AM-HM} \\ \\ \mbox{Using AM of mth power}\leq\mbox{mth power of AM for} \ m<1 \ \mbox{we have} \\ \\ \frac{\sum \sqrt{1+\frac{8bc}{a^2}}}{3}\leq (\frac{3+\sum \frac{8bc}{a^2}}{3})^{\frac{1}{2}} \\ \\ \sum \frac{8bc}{a^2} \geq \frac{9}{\sum \frac{a^2}{8bc}} \\ \\ \mbox{Now the above denominator} \leq \frac{3}{8} \ \mbox{using Rearrangement inequality} \\ \\ \mbox{Thus} \ \sum \frac{8bc}{a^2} \geq 24 \\ \\ \mbox{Thus} \ \sum \frac{1}{\sqrt{1+\frac{8bc}{a^2}}}\geq \frac{9}{9} = 1$

isnt this imo problem????

why isnt it in olympiad section??

Anyway the idea is to use holder's inequality:

(1)computerboosy,

you are wrong here.

Now the above denominator < 3/8 using Rearrangement inequality.

Either i dont follow or i am being stupid…..

hell ever this is what i meant …

$\sum \frac{a^2}{8bc} \geq \frac{(a+b+c)^2}{8(ab+bc+ca)} \\ \\ \mbox{The above step follows from} \ \frac{x^2}{a}+\frac{y^2}{b} \geq \frac{(x+y)^2}{a+b} \\ \\ \mbox{Now} \ \frac{(a+b+c)^2}{8(ab+bc+ca)} = \frac{a^2+b^2+c^2+2(ab+bc+ca)}{8(ab+bc+ca)} \\ \\ \mbox{Now using Rearrangement inequality we have} \ \frac{(a+b+c)^2}{8(ab+bc+ca)} \leq \frac{3}{8}$

In coherence with Computerboosy's solution,

here is the titu's lemma: (you might have heard of it). Its extremely poweful.

It is:

$\frac{x^{2}}{a} + \frac{y^{2}}{b} + \frac{z^{2}}{c} ... >= \frac{(x+y+z...)^{2}}{a+b+c...}$

This result can be easily proved for two terms and then extended further by an inductive argument.

In fact this gives us another proof of the very famous nesbitt inequality.

Here are a couple of problems on this spirit:

Problem 1:

$\Sigma \frac{1}{a^{3}(b+c)} >= 1.5$ when

$a,b,c$ are positive reals with product 1.

Problem 2:

If $a,b,c$be positive reals such that

then prove that

(2)computerboosy, u r wrong again here.

(1)that is obviously wrong.

aur rearrangement inequality ka toh savaal hi nahi uthta

and raghav, are u sure its titu's lemma??

I believe its engel form of cauchy…