11From Tchebycheff, \sum{\frac{a}{4b^2+1}}\ge \frac{3(a+b+c)}{4(a^2+b^2+c^2)+3}
I guess, CS on Denominator does the rest.
\texttt{let a,b,c be positive reals} \\ \texttt{if} \ a+b+c=3 \\ \texttt{then prove that:}\\ \frac{a}{4b^{2}+1}+\frac{b}{4c^{2}+1}+\frac{c}{4a^{2}+1}\geq\frac{3}{5}
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9 Answers
11
1hey but are conditions for tchebychef's inequality satisfied ???
i dont think so
341@soumik: i think you are equating cyclic expression with symmetric.
1i really dont know what kind of inequality is soumik applying
chebyshef inequality is not that i guess
11@ qsn setter, the inequality that I used was ΣA1a1 ≥n(ΣAi)Σ(ai), provided ai≥ai+1.
Directly obtained from Chebucheff. [56]
341that is another story altogether :D. I am wary of death traps like that!
11Yup. that's exactly what I observed after #5.
In my earlier post, I just stated how I reached that inequality from chebycheff.