Neocijenjeno
April 15, 2025, 1:45 p.m. (1 week)
Dokažite da za pozitivne realne brojeve
i
vrijedi nejednakost
![\sqrt[3]{\frac ab} + \sqrt[3]{\frac ba} \leqslant \sqrt[3]{2\left(a+b\right)\left(\frac 1a + \frac 1b\right)} \text{.}](/media/m/4/6/2/462032d0c948d6c777de953290b33bd0.png)
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Dokažite da za pozitivne realne brojeve $a$ i $b$ vrijedi nejednakost $$\sqrt[3]{\frac ab} + \sqrt[3]{\frac ba} \leqslant \sqrt[3]{2\left(a+b\right)\left(\frac 1a + \frac 1b\right)} \text{.}$$
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Iz Hölderove nejednakosti imamo ![\sqrt[3]{2(a+b)\left(\frac{1}{a} + \frac{1}{b}\right)} = (1+1)^{\frac{1}{3}}\cdot(a + b)^{\frac{1}{3}} \cdot \left(\frac{1}{a}+ \frac{1}{b} \right)^{\frac{1}{3}} \geq \sqrt[3]{\frac{a}{b}} + \sqrt[3]{\frac{b}{a}}](/media/m/f/1/2/f12cbc4a8f11449ec059a683a20ce063.png)
Iz Hölderove nejednakosti imamo
$$
\sqrt[3]{2(a+b)\left(\frac{1}{a} + \frac{1}{b}\right)} = (1+1)^{\frac{1}{3}}\cdot(a + b)^{\frac{1}{3}} \cdot \left(\frac{1}{a}+ \frac{1}{b} \right)^{\frac{1}{3}} \geq \sqrt[3]{\frac{a}{b}} + \sqrt[3]{\frac{b}{a}}
$$
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