IIT JEE , AIEEE Forum - Mathematics , Algebra

sandeepramesh

OMG! can someone plz latexify it for me? Thanks

konichiwa2x

sandeepramesh's post:

Let me prove a genl inequality first whose's case is this sum:
To prove

let

Applying Weierstress's inequality,

But

On simplification you get

from 1 and 2

Hence on simplification we get the result Smile

sandeepramesh

thanx but i found a soln to that problem Mr. Green

konichiwa2x

lol

you should have told...by the way nice solution...

konichiwa2x

Ok here is another one.

2. Suppose $a,b,c$ are positive reals such that $abc=1$ , show:
$\frac{1}{\sqrt{(1+b)(1+c)}}+\frac{1}{\sqrt{(1+c)(1+a)}}+\frac{1}{\sqrt{(1+a)(1+b)}}\leq \frac{3}{2}$

sandeepramesh

well i typed the thing on another place and just copied Mr. Green

just like these wonderful smileys Mr. Green

they are nice no? i love them.

sandeepramesh

very easy

typing the soln wait

konichiwa2x

yeah definitely better smileys..where from? aops?

sandeepramesh

Put $a = \frac {x}{y}, b = \frac {y}{z}, a = \frac {z}{x}$
The inequality reduces to:
$\sum \frac {\sqrt{x \cdot z}}{\sqrt{(x + y) \cdot (y + z)}} \leq \frac {3}{2} \implies \sum \frac {\sqrt{x \cdot z \cdot (x + z)}}{\sqrt{(x + y) \cdot (y + z) \cdot (z + x)}} \leq \frac {3}{2}$
This is a simple follow through of the AM GM inequality Mr. Green

Yes smileys from aops. You there in aops?

konichiwa2x

lol.. nice dude.. Your method is better than what I had in mind.

Let $a=x^{3},b=y^{3},c=z^{3}$ , then

$\sum\frac{1}{\sqrt{(1+b)(1+c)}}=\sum\frac{xyz}{\sqrt{(xyz+y^{3})(xyz+z^{3})}}$

$=\sum x\sqrt{\frac{yz}{(y^{2}+zx)(z^{2}+xy)}}\leq\sum\frac{10x+y+z}{8(x+y+z)}=\frac{3}{2}$ .

Proof:

sandeepramesh

btw what are you man? A student or a teacher? Mr. Green

konichiwa2x

no i am not "there" in aops as such..But I do look at the problems discussed there often.

konichiwa2x

I am a student! What made u think I am a teacher?

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