Genelleştirilmiş JBMO 2002 #4 {çözüldü}

[Hüseyin Yiğit EMEKÇİ]

$a,b,c$ pozitif reeller olmak üzere

$$\sum_{n=1}^{2k+1}{\dfrac{1}{a_{n}\left(a_{n-2k+1}+a_{n-2k+2}+\cdots+a_{n}\right)}}\geq \dfrac{(2k+1)^3}{2k\left(\sum_{cyc}{a_{k}}\right)}$$

olduğunu gösteriniz.

[Hüseyin Yiğit EMEKÇİ]

Toplam işaretini dairesel eşitsizlik hâline formatına getirelim

$$\sum_{cyc}{\dfrac{1}{a_{2k}(a_{1}+a_{2}+\cdots+a_{2k})}}\geq p$$

$$\rightarrow S=\left(\sum_{cyc}{\dfrac{1}{a_{2k}(a_{1}+a_{2}+\cdots+a_{2k})}}\right)\left(\sum_{cyc}{a_{1}}\right)\left(\sum_{cyc}{\left(a_{1}+a_{2}+\cdots+a_{2k}\right)}\right)\geq p\left(\sum_{cyc}{a_{1}}\right)\left(\sum_{cyc}{\left(a_{1}+a_{2}+\cdots+a_{2k}\right)}\right)$$

Sol tarafı çözümleyeceğiz

$i)$
$$\left(\sum_{cyc}{\dfrac{1}{a_{2k}(a_{1}+a_{2}+\cdots+a_{2k})}}\right)\overbrace{\geq}^{AGO} (2k+1).\sqrt[2k+1]{\dfrac{1}{\left(\prod{a_{1}}\right)\left(\prod{a_{1}+a_{2}+\cdots+a_{2k}}\right)}}$$

$ii)$
$$\left(\sum_{cyc}{a_{1}}\right)\overbrace{\geq}^{AGO} (2k+1).\sqrt[2k+1]{\prod{a_{1}}}$$

$iii)$
$$\left(\sum_{cyc}{\left(a_{1}+a_{2}+\cdots+a_{2k}\right)}\right)\overbrace{\geq}^{AGO} (2k+1).\sqrt[2k+1]{\prod{a_{1}+a_{2}+\cdots+a_{2k}}}$$

Dolayısıyla
$$S\geq (2k+1)^3.\sqrt[2k+1]{\left(\dfrac{1}{\left(\prod{a_{1}}\right)\left(\prod{a_{1}+a_{2}+\cdots+a_{2k}}\right)}\right)\left(\prod{a_{1}}\right)\left(\prod{a_{1}+a_{2}+\cdots+a_{2k}}\right)}$$
$$=(2k+1)^3$$

$$S\geq (2k+1)^3\geq p\left(\sum_{cyc}{a_{1}}\right)\left(\sum_{cyc}{\left(a_{1}+a_{2}+\cdots+a_{2k}\right)}\right)$$
$$\rightarrow p\leq \dfrac{(2k+1)^3}{\left(\sum_{cyc}{a_{1}}\right)\left(\sum_{cyc}{\left(a_{1}+a_{2}+\cdots+a_{2k}\right)}\right)}=\dfrac{(2k+1)^3}{\left(\sum_{cyc}{a_{1}}\right).2k\left(\sum_{cyc}{a_{1}}\right)}=\dfrac{(2k+1)^3}{2k\left(\sum_{cyc}{a_{1}}\right)^2}$$

Sonuç olarak
$$\sum_{cyc}{\dfrac{1}{a_{2k}(a_{1}+a_{2}+\cdots+a_{2k})}}\geq \dfrac{(2k+1)^3}{2k\left(\sum_{cyc}{a_{1}}\right)^2}\geq p$$

İspat biter.