let $a_{1},a_{2},\cdots,a_{n}>0$,and such $$a_{1}a_{2}\cdot \cdots a_{n}=1$$

prove or disprove: $$\dfrac{1}{a^5_{1}(a_{2}+2a_{3})^2}+\dfrac{1}{a^5_{2}(a_{3}+2a_{4})^2}+\cdots+\dfrac{1}{a^5_{n}(a_{1}+2a_{2})^2}\ge\dfrac{n}{9}$$

when $n=3$ is [USA TST 2010 2] http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1960090

and I post this proof:

since $a_{1}a_{2}a_{3}=1$,then $$a_{1}\to \dfrac{1}{a},a_{2}\to \dfrac{1}{b},a_{3}\to\dfrac{1}{c},abc=1$$ $$\dfrac{1}{a^5_{1}(a_{2}+2a_{3})^2}=\dfrac{a^5b^2c^2}{(c+2b)^2}=\dfrac{a^3}{(c+2b)^2}$$ so Use Holder inequality we have $$\left(\sum_{cyc}\dfrac{a^3}{(c+2b)^2}\right)\left(\sum_{cyc}(c+2b)\right)\left(\sum_{cyc}(c+2b)\right)\ge\left(a+b+c\right)^3$$ so $$\sum_{cyc}\dfrac{a^3}{(c+2b)^2}\ge \dfrac{a+b+c}{9}\ge\dfrac{3\sqrt[3]{abc}}{9}=\dfrac{1}{3}$$

But for $n$ I can't prove it,Thank you

because I use $$a_{i}\to\dfrac{1}{a_{i}},a_{1}a_{2}\cdots a_{n}=1$$ then $$\dfrac{1}{a^5_{1}(a_{2}+2a_{3})^2}=\dfrac{a^5_{1}a^2_{2}a^2_{3}}{(a_{3}+2a_{2})^2}\neq\dfrac{a^3_{1}}{(a_{3}+2a_{2})^2}$$ because $a_{1}a_{2}a_{3}\neq 1$,in fact $a_{1}a_{2}\cdots a_{n}=1$

so I can't use Holder inequality, and then How prove it?

Thank you very much。

  • $\begingroup$ By the Purkiss Principle ( from maa.org/programs/maa-awards/writing-awards/… ) we have that the left hand side is minimal $= n/9$ , for : $a_1=a_2=\cdots =a_n=1$ . But then we have to prove (again and again) that this local minimum is a global minimum, which is .. a bit harder :-( $\endgroup$ – Han de Bruijn Feb 11 '14 at 20:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy