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Wildan Bagus W

Nilai minimum

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Untuk $a,b,c > 0$ dan $abc = 1$, tentukan nilai minimum dari

\[ \frac{b+c}{\sqrt{a}} + \frac{a+c}{\sqrt{b}} + \frac{a+b}{\sqrt{c}}\]

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$a, b, c>0,\, abc=1$. Dengan ketaksamaan AM-GM dua kali, maka \begin{align*}\frac{b+c}{\sqrt{a}}+\frac{a+c}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}&\geq\frac{2\sqrt{bc}}{\sqrt{a}}+\frac{2\sqrt{ac}}{\sqrt{b}}+\frac{2\sqrt{ab}}{\sqrt{c}}\\&=2\left(\sqrt{\frac{bc}{a}}+\sqrt{\frac{ac}{b}}+\sqrt{\frac{ab}{c}}\right)\\&\geq2\cdot3\cdot\sqrt[3]{\sqrt{\frac{bc}{a}}\sqrt{\frac{ac}{b}}\sqrt{\frac{ab}{c}}}\\&=6\cdot\sqrt[3]{\frac{abc}{\sqrt{abc}}}\\&=\boxed{6}.\end{align*} Jadi, nilai minimumnya $6$.

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\[ \frac {b+c}{\sqrt a} + \frac{c+a}{\sqrt b} + \frac{a+b}{\sqrt c} = \frac{b}{\sqrt a} + \frac{c}{\sqrt a} + \frac{c}{\sqrt b} + \frac{a}{\sqrt b} + \frac{a}{\sqrt c}+\frac{b}{\sqrt c} = 6 \sqrt[6]{\frac{a^2b^2c^2}{abc}} = 6 \]

Edited by OnePunchMan

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