Školjka

$\frac{a}{a+b^2} = 1 - \frac{b^2}{a+b^2}$ pa je nejednakost ekvivalentna sa: $\frac{1}{4}(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) + \frac{b^2}{a+b^2} + \frac{c^2}{b+c^2} + \frac{a^2}{c+a^2} \geq 3$ AG $\frac{b^2}{a+b^2} + \frac{9}{16}(a+b^2) \geq \frac{3}{2}b \iff \frac{b^2}{a+b^2} \geq \frac{3}{2}b - \frac{9}{16}a - \frac{9}{16}b^2$ Slijedi $\frac{1}{4}(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) + \frac{b^2}{a+b^2} + \frac{c^2}{b+c^2} + \frac{a^2}{c+a^2} \geq$ $ \frac{1}{4}(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) + \frac{3}{2}(a+b+c) - \frac{9}{16}(a+b+c) - \frac{9}{16}(a^2+b^2+c^2) = $ $ \frac{1}{4}(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})+ \frac{15}{16} - \frac{9}{16}(a^2+b^2+c^2) $ Dovoljno je dokazati $ \frac{1}{4}(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})+ \frac{15}{16} - \frac{9}{16}(a^2+b^2+c^2) \geq 3 \iff$ $ 4(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \geq 33 + 9(a^2+b^2+c^2) = 33 + 9(a+b+c)^2 - 18(ab+bc+ca) \iff$ $2(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) + 9ab +9bc +9ca \geq 21$ $\iff$ $2(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) + 9ab +9bc +9ca + 3 + 12(a+b+c) \geq 36$ AG $9ab +1 \geq 6\sqrt{ab}$ $9bc +1 \geq 6\sqrt{bc}$ $9ca +1 \geq 6 \sqrt{ca}$ $6a + 6b \geq 12\sqrt{ab}$ $6b + 6c \geq 12\sqrt{bc}$ $6c + 6a \geq 12\sqrt{ca}$ Slijedi $2(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) + 9ab +9bc +9ca + 3 + 12(a+b+c) \geq 2(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) + 18\sqrt{ab} +18\sqrt{bc} +18\sqrt{ca}$ AG $\frac{1}{a} + \frac{1}{b} + 9\sqrt{ab} + 9\sqrt{ab} \geq 4 \cdot 81^{\frac{1}{4}} = 12$ Zbrajanjem za ostale članove slijedi $2(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) + 18\sqrt{ab} +18\sqrt{bc} +18\sqrt{ca} \geq 36$ QED