INEQUALITY'S BLOG

August 1, 2010

Problem 031

Filed under: Uncategorized — KKKVVV @ 12:44 am

Let a\geq b\geq c>0 and \displaystyle 3a^2\leq abc ; 3b^2\leq abc ; 3c^2\leq abc. Prove that:

\displaystyle \frac{a^2b^2}{a+bc}+\frac{b^2c^2}{b+ca}+\frac{c^2a^2}{c+ab}\geq abc\biggl(\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{c}{c+ab}\biggl)

 

Van Khea     

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