Školjka

Primijetimo da \begin{center} $8(a+b+c)=\sum_{}\frac{1}{a^2}+7b=\sum_{}\frac{7a^2b+1}{a^2}$\\ \end{center} Dakle, po Hölderu imamo:\begin{center} $(2(a+b+c))^3=\sum_{}\frac{7a^2b+1}{a^2}(a+b+c)(a+b+c) \geq (\sum_{}\sqrt[3]{7a^2b+1})^3$\\\\ $\Leftrightarrow (2(a+b+c))^3 \geq (\sum_{}\sqrt[3]{7a^2b+1})^3$\\\\ $\Leftrightarrow 2(a+b+c) \geq \sqrt[3]{7a^2b+1}+\sqrt[3]{7b^2c+1}+\sqrt[3]{7c^2a+1}$ \end{center}