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Fantezi Cebir => Analiz-Cebir => Konuyu başlatan: Hüseyin Yiğit EMEKÇİ - Eylül 24, 2023, 07:10:34 ös
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$a,b,c$ pozitif reeller olmak üzere
$$\sum_{cyc}{\dfrac{\left((k+1)a+kb+kc\right)^2}{(k+1)a^2+k(b+c)^2}}\leq 6k+\dfrac{3}{2k+1}+1$$
olduğunu gösteriniz.
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Eşitsizliğin homojen özellikte olduğuna dikkat edelim. O zaman $a+b+c=1$ olsun.
$$\sum_{cyc}{\dfrac{\left((k+1)a+kb+kc\right)^2}{(k+1)a^2+k(b+c)^2}}=\sum_{cyc}{\dfrac{(a+k)^2}{(k+1)a^2+k(1-a)^2}}=\sum_{cyc}{\dfrac{(a+k)^2}{(2k+1)a^2-2ak+k}}$$
$$=\sum_{cyc}{\left(\dfrac{1}{2k+1}+\dfrac{2ak+\dfrac{2ak}{2k+1}+k^2-\dfrac{k}{2k+1}}{(2k+1)a^2-2ak+k}\right)}=S$$
Paydadaki ifadeye bir eşitsizlik hamlesinde bulunalım.
$$(2k+1)a^2-2ak+k=(2k+1)a^2-2ak+\dfrac{k^2}{2k+1}+k-\dfrac{k^2}{2k+1}\overbrace{\geq}^{AGO} 2ak-2ak+k-\dfrac{k^2}{2k+1}=k-\dfrac{k^2}{2k+1}$$
$$S\leq \sum_{cyc}{\left(\dfrac{1}{2k+1}+\dfrac{2ak+\dfrac{2ak}{2k+1}+k^2-\dfrac{k}{2k+1}}{k-\dfrac{k^2}{2k+1}}\right)}=\dfrac{3}{2k+1}+\dfrac{2k(a+b+c)+\dfrac{2k}{2k+1}(a+b+c)+3k^2-\dfrac{3k}{2k+1}}{k-\dfrac{k^2}{2k+1}}$$
$$=\dfrac{3}{2k+1}+\dfrac{2k+\dfrac{2k}{2k+1}+3k^2-\dfrac{3k}{2k+1}}{k-\dfrac{k^2}{2k+1}}=\dfrac{3}{2k+1}+\dfrac{(2k+1)\left(3k^2+2k-\dfrac{k}{2k+1}\right)}{k^2+k}$$
$$=\dfrac{3}{2k+1}+\dfrac{(2k+1)k(3k+2-\dfrac{1}{2k+1}}{k(k+1)}=\dfrac{3}{2k+1}+\dfrac{(2k+1)(3k+2-\dfrac{1}{2k+1})}{k+1}=\dfrac{3}{2k+1}+\dfrac{(2k+1)\left(\dfrac{6k^2+7k+1}{2k+1}\right)}{k+1}$$
$$=\dfrac{3}{2k+1}+\dfrac{6k^2+7k+1}{k+1}=\dfrac{3}{2k+1}+6k+1$$
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Genelleştirme 2
$a_{1},a_{2},\cdots,a_{n}$ pozitif reeller olmak üzere
$$\sum_{cyc}{\dfrac{\left((k+1)a_{1}+k(a_{2}+a_{3}+\cdots+a_{n}\right)^2}{(k+1)a_{1}^2+k(a_{2}+a_{3}+\cdots+a_{n})}}\geq n(2k-1)+\dfrac{n}{2k+1}+4$$
olduğunu gösteriniz.
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Genel problemin çözümü yine homojeniteden dolayı $\sum_{cyc}{a_{1}}=1$ atamasıyla başlıyor. Cebirsel manipülasyonlar sonucu
$$\leq \sum_{cyc}{\left(\dfrac{1}{2k+1}+\dfrac{2a_{1}k+\dfrac{2a_{1}k}{2k+1}+k^2-\dfrac{k}{2k+1}}{k-\dfrac{k^2}{2k+1}}\right)}=\dfrac{n}{2k+1}+\dfrac{2k(\sum_{cyc}{a_{1}})+\dfrac{2k}{2k+1}(\sum_{cyc}{a_{1}})+nk^2-\dfrac{nk}{2k+1}}{k-\dfrac{k^2}{2k+1}}$$
Sondaki ifadenin ispatı orijinal sorunun çözümînde yapılmıştır.
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$$\dfrac{n}{2k+1}+\dfrac{2k+\dfrac{2k}{2k+1}+nk^2-\dfrac{nk}{2k+1}}{k-\dfrac{k^2}{2k+1}}=\dfrac{n}{2k+1}+\dfrac{\dfrac{k(2-n)}{2k+1}+k(2+nk)}{k-\dfrac{k^2}{2k+1}}$$
$$=\dfrac{n}{2k+1}+\dfrac{\dfrac{2-n}{2k+1}+2+nk}{\dfrac{k+1}{2k+1}}=\dfrac{n}{2k+1}+\dfrac{2nk^2+nk-n+4k+4}{k+1}$$
$$=\dfrac{n}{2k+1}+\dfrac{n(2k^2+k-1)}{k+1}+4=\dfrac{n}{2k+1}+\dfrac{n(2k-1)(k+1)}{k+1}+4$$
$$=\dfrac{n}{2k+1}+n(2k-1)+\dfrac{n}{2k+1}+4$$
olduğundan dolayı eşitsizlik çalışır.