Školjka

%V0 Promatrajmo funkciju $f(x)=x^{\frac{1}{3}}$ odakle proizlazi $f''(x)=-\frac{2}{9}x^{-\frac{5}{3}}\leq 0$ dakle možemo po Jensenovj nejednakosti zaključiti da vrijedi $$\sum_{cyc} f\bigg(\frac{1}{a}+6b\bigg) \leq 3 f\bigg(\frac{1}{3}\big(\sum_{cyc} \frac{1}{a} + \sum_{cyc} 6a \big)\bigg)$$ dakle ostaje nam dokazati da je $$\sqrt[3]{9\frac{1}{abc} +54\sum_{cyc}a}\leq \frac{1}{abc}$$ $$\iff 9a^2b^2c^2 +54\sum_{cyc}a^4b^3c^3\leq 1$$ Što dovršavama vrlo jednostavno sa dvije nejednakosti $$ 3\sqrt[3]{a^2b^2c^2} \leq ab+bc+ca =1 \implies a^2b^2c^2\leq\frac{1}{27}$$ $$abc(a+b+c)\leq\frac{(ab+bc+ca)^2}{3}=\frac{1}{3}$$ Dakle $$9a^2b^2c^2 +54\sum_{cyc}a^4b^3c^3=a^2b^2c^2 \big( 9 + 54 abc ( a + b+ c) \big) \leq \frac{1}{27}\big( 9 +54 \frac{1}{3}\big)=1$$