Školjka

Koristeći rearrangement inequality (ili Muirhead, kako hoćete) imamo $$a^5+b^5\ge a^4b+b^4a\Longleftrightarrow a^5+b^5+ab\ge a^4b+b^4a+ab\Longleftrightarrow \frac{ab}{a^5+b^5+ab}\le\frac{ab}{a^4b+b^4a+ab}=\frac{1}{a^3+b^3+1}$$ pa analogno imamo $\frac{bc}{bc+b^5+c^5}\le \frac{1}{b^3+c^3+1}$ $\frac{ac}{ac+a^5+c^5}\le \frac{1}{a^3+c^3+1}$ nadalje, isto koristeći rearrangemet (ili Muirhead) imamo (koristeći $abc=1$) $$a^3+b^3\ge a^2b+b^2a\Longleftrightarrow a^3+b^3+1\ge a^2b+b^2a+1\Longleftrightarrow \frac{1}{a^3+b^3+1}\le\frac{1}{a^2b+b^2a+1}=\frac{abc}{a^2b+b^2a+abc}=\frac{c}{a+b+c}$$ te analogno imamo $\frac{1}{a^3+c^3+1}\le \frac{b}{a+b+c}$ $\frac{1}{b^3+c^3+1}\le \frac{a}{a+b+c}$ korištenjem svih prethodnih nejednakosti na kraju imamo $$\sum \frac{ab}{ab+a^5+b^5}\le \sum \frac{1}{a^3+b^3+1}\le\sum\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1$$