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Fantezi Cebir => Analiz-Cebir => Konuyu başlatan: Hüseyin Yiğit EMEKÇİ - Kasım 05, 2023, 06:31:36 ös
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Genelleştirme 1
$a,b,c,\lambda_i$ pozitif reeller olmak üzere
$$\left(\lambda_1 +\dfrac{a}{b}\right)\left(\lambda_2 +\dfrac{b}{c}\right)\left(\lambda_3 +\dfrac{c}{a}\right)\geq 2\left(\dfrac{a\sqrt[3]{\lambda_2\lambda_3^2}+b\sqrt[3]{\lambda_3\lambda_1^2}+c\sqrt[3]{\lambda_1\lambda_2^2}}{\sqrt[3]{abc}}+2\lambda_1\lambda_2\lambda_3-1\right)$$
olduğunu gösteriniz.
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$$\lambda =2,k=p=1$$
verilirse problem APMO 1998 #3 (https://geomania.org/forum/index.php?topic=8761.msg23952;topicseen#new) 'e dönüşür.
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Verilen ifadeyi açarsak
$$LHS=\left(\lambda_1 +\dfrac{a}{b}\right)\left(\lambda_2 +\dfrac{b}{c}\right)\left(\lambda_3 +\dfrac{c}{a}\right)= \left(\lambda_1\lambda_2+\lambda_1\dfrac{b}{c}+\lambda_2\dfrac{a}{b}+\dfrac{a}{c}\right)\left(\lambda_3+\dfrac{c}{a}\right)$$
$$=\lambda_1\lambda_2\dfrac{c}{a}+\lambda_2 \dfrac{c}{b}+\lambda_2\lambda_3\dfrac{a}{b}+\lambda_3\dfrac{a}{c}+\lambda_1\lambda_3\dfrac{b}{c}+\lambda_1\dfrac{b}{a}+\lambda_1\lambda_2\lambda_3+1=S_a+S_b+S_c+\lambda_1\lambda_2\lambda_3+1$$
Şimdi $S_a=\lambda_2\lambda_3\dfrac{a}{b}+\lambda_3\dfrac{a}{c}$ der ve $S_b$ ve $S_c$ 'yi de aynı şekilde yukarıdaki gibi tanımlarsak ve sonrasında Aritmetik-Geometrik Ortalama kullanırsak
$$S_a+1=\lambda_2\lambda_3\dfrac{a}{b}+\lambda_3\dfrac{a}{c}+1\overbrace{\geq}^{AGO} 3\sqrt[3]{\dfrac{\lambda_2\lambda_3^2a^2}{bc}}=\dfrac{3a\sqrt[3]{\lambda_2\lambda_3^2}}{\sqrt[3]{abc}}$$
$$\Rightarrow $S_a\geq \dfrac{3a\sqrt[3]{\lambda_2\lambda_3^2}}{\sqrt[3]{abc}}-1$$
Diğer $S_b$ ve $S_c$ için de benzer
$$S_b\geq \dfrac{3b\sqrt[3]{\lambda_3\lambda_1^2}}{\sqrt[3]{abc}}-1 \qquad \text{ve} \qquad S_c\geq \dfrac{3c\sqrt[3]{\lambda_1\lambda_2^2}}{\sqrt[3]{abc}}-1$$
ifadeleri rahatlıkla elde edilebilir. Buna göre
$$LHS=S_a+S_b+S_c+\lambda_1\lambda_2\lambda_3+1\geq \dfrac{3a\sqrt[3]{\lambda_2\lambda_3^2}}{\sqrt[3]{abc}}+\dfrac{3b\sqrt[3]{\lambda_3\lambda_1^2}}{\sqrt[3]{abc}}+\dfrac{3c\sqrt[3]{\lambda_1\lambda_2^2}}{\sqrt[3]{abc}}+\lambda_1\lambda_2\lambda_3-2\geq 2\left(\dfrac{a\sqrt[3]{\lambda_2\lambda_3^2}+b\sqrt[3]{\lambda_3\lambda_1^2}+c\sqrt[3]{\lambda_1\lambda_2^2}}{\sqrt[3]{abc}}+2\lambda_1\lambda_2\lambda_3-1\right)$$
sondaki eşitsizliği göstermemiz ispatı sonlandıracaktır. Gösterelim. Biraz düzenlersek
$$ \dfrac{3a\sqrt[3]{\lambda_2\lambda_3^2}}{\sqrt[3]{abc}}+\dfrac{3b\sqrt[3]{\lambda_3\lambda_1^2}}{\sqrt[3]{abc}}+\dfrac{3c\sqrt[3]{\lambda_1\lambda_2^2}}{\sqrt[3]{abc}}+\lambda_1\lambda_2\lambda_3\geq 2\left(\dfrac{a\sqrt[3]{\lambda_2\lambda_3^2}+b\sqrt[3]{\lambda_3\lambda_1^2}+c\sqrt[3]{\lambda_1\lambda_2^2}}{\sqrt[3]{abc}}+2\lambda_1\lambda_2\lambda_3\right)$$
$$\Rightarrow \dfrac{a\sqrt[3]{\lambda_2\lambda_3^2}+b\sqrt[3]{\lambda_3\lambda_1^2}+c\sqrt[3]{\lambda_1\lambda_2^2}}{\sqrt[3]{abc}}\geq 3\lambda_1\lambda_2\lambda_3$$
elde ederiz. Bu ifadeyi ise Aritmetik-Geometrik Ortalama'dan açık bir şekilde
$$\dfrac{a\sqrt[3]{\lambda_2\lambda_3^2}+b\sqrt[3]{\lambda_3\lambda_1^2}+c\sqrt[3]{\lambda_1\lambda_2^2}}{\sqrt[3]{abc}}\geq \dfrac{3\sqrt[3]{\lambda_1^3\lambda_2^3\lambda_3^3abc}}{\sqrt[3]{abc}}=3\lambda_1\lambda_2\lambda_3$$
elde eder vs ispatı tamamlarız.