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Fantezi Cebir => Analiz-Cebir => Konuyu başlatan: Hüseyin Yiğit EMEKÇİ - Ekim 08, 2023, 10:57:00 ös
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$a,b,c$ 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.
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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.