If $a,b,c>0$ and $abc=3$,find the biggest value of: $\frac{a^2b^2}{a^7+a^3b^3c+b^7}+\frac{b^2c^2}{b^7+b^3c^3a+c^7}+\frac{c^2a^2}{c^7+c^3a^3b+a^7}$
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Tags: inequalities, inequalities unsolved
18.07.2011 20:31
hello, with the help of the method of Lagrange Multipliers i have no maximum found. Sonnhard.
18.07.2011 22:31
Dr. Sonhard, i think you´re wrong. We are going tu use the next well known inequality $a^7+b^7\geq a^3b^3(a+b)$ So we have $\frac{a^2b^2}{a^7+a^3b^3c+b^7}\leq \frac{a^2b^2}{a^3b^3(a+b)+a^3b^3c}=\frac{c}{3(a+b+c)}$ By a similar argument, we have $\sum_{cyc}\frac{a^2b^2}{a^7+a^3b^3c+b^7}\leq \sum_{cyc}\frac{c}{3(a+b+c)}=\frac{1}{3}$ We attain the maximum when $a=b=c=3^{\frac{1}{3}}$
18.07.2011 22:54
hello, thank you for your nice solution,i have only said that i have found no such maximum with the method of the Lagrange Multipliers, this means not that there is no such maximum. Thank you again for your nice solution. Sonnhard.