Cho $a, b, c >0$ và $a+b+c+ab+ac+bc=6abc$ chứng minh rằng $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \geq 3$
Edited by Ngoc Hung, 23-03-2023 - 06:10.
Chứng Minh Rằng $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \geq 3$
Started By nguyetnguyet829, 16-03-2023 - 12:27
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Best Answer Ngoc Hung, 23-03-2023 - 06:08
Ta có $\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6$
$\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\geq \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}$
$\frac{1}{a^{2}}+1\geq \frac{2}{a};\frac{1}{b^{2}}+1\geq \frac{2}{b};\frac{1}{c^{2}}+1\geq \frac{2}{c}$
Suy ra $$3\left ( \frac{1}{a^{2}} +\frac{1}{b^{2}}+\frac{1}{c^{2}}\right )+3\geq 2\left ( \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )=12\Leftrightarrow \frac{1}{a^{2}} +\frac{1}{b^{2}}+\frac{1}{c^{2}}\geqslant 3$$
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Posted 23-03-2023 - 06:08
Ngoc Hung
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Ta có $\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6$
$\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\geq \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}$
$\frac{1}{a^{2}}+1\geq \frac{2}{a};\frac{1}{b^{2}}+1\geq \frac{2}{b};\frac{1}{c^{2}}+1\geq \frac{2}{c}$
Suy ra $$3\left ( \frac{1}{a^{2}} +\frac{1}{b^{2}}+\frac{1}{c^{2}}\right )+3\geq 2\left ( \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )=12\Leftrightarrow \frac{1}{a^{2}} +\frac{1}{b^{2}}+\frac{1}{c^{2}}\geqslant 3$$
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