SinaQane wrote: For some positive numbers $a$, $b$, $c$ and $d$, we know that $$ \frac{1}{a^3 + 1}+ \frac{1}{b^3 + 1}+ \frac{1}{c^3 + 1} + \frac{1}{d^3 + 1} = 2. $$Prove that $$ \frac{1 - a}{a^2 - a + 1} + \frac{1-b}{b^2 - b + 1} + \frac{1-c}{c^2 - c + 1} +\frac{1-d}{d^2 - d + 1} \geq 0. $$ Because \[\frac{1-a}{a^2-a+1}-\frac 43\left(\frac{1}{a^3+1} - \frac 12\right) = \frac{(2a+1)(a-1)^2}{3(a+1)(a^2-a+1)} \geqslant 0.\]