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[Post New]14 Mar 2008 13:17:43 IST
    Subject: Re:An inequality
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hsbhatt (4460)

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\text{I must confess I am in love with Rearrangement Inequality. Hence this proof} \\ \\ \text {We have seen in the previous post of mine that} \\ \\ \text {If we assume} \ a \geq b \geq c \ \text{then} \\ \\ \frac {a^2} {b^2+c^2} \geq \frac {b^2} {c^2+a^2} \geq \frac {c^2} {a^2+b^2} \\ \\ \text{If we call the given expression as I, then from RI, we have} \\ \\ I \geq \sum \frac {a b^2} {b^2+c^2} \ \text {and also} \\ \\ I \geq \sum \frac {a c^2} {b^2+c^2} \\ \\ \text {adding the above two} \\ \\ 2I \geq (a+b+c) \\ \\ \text {Hence} I \geq \frac{1} {2}
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