21is this rite?
I think many of the JEE aspirants fear inequalities a lot so a very simple one. It uses only AM GM inequality That too only once!
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq 3/2
Just give it a try .... Think for just a bit and all of a small tricks that you can apply. it is really a simple one.
a,b,c > 0
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21#### galat hai #####
First cuple of steps havnt bn shown but i guesss they are self-Understandable [1]
1\text{Consider} \sum_{cyc}(\frac{a}{b+c}) \\ \text{Apply AM GM} \\ \\ \Longrightarrow \sum_{cyc}(\frac{a}{b+c}) \ge 3(\frac{abc}{(a+b)(b+c)(c+a)})^{1/3} \\Now,(a+b)(b+c)(c+a) \ge 8(abc) \\ \Longrightarrow \sum_{cyc}(\frac{a}{b+c}) \ge 3(\frac{abc}{8abc})^{1/3} \\ \Longrightarrow \sum_{cyc}(\frac{a}{b+c}) \ge \frac{3}{2}
1i'v used AM-HM inequality...
(a+b)+(a+c)+(b+c) ≥ 3
--------------------------- ------------------------------------------
3 1/(a+b) + 1/(a+c) +1/(b+c)
2(a+b+c)( 1/(a+b) + 1/(a+c) +1/(b+c) ) ≥ 9
c/(a+b) + b/(a+c) +a/(b+c) ≥ 9/2 - 3
c/(a+b) + b/(a+c) +a/(b+c) ≥ 3/2.
is dis correct?
66This is the famous Nesbitt's Inequality. Look at the following link for as many as 25 proofs. (Go to the link, and look for the link on Nesbitt's)
http://kaymant.googlepages.com/mathematics