Let $\sum\frac{1}{a^3+1}=2$. Prove that $\sum\frac{1-a}{a^2-a+1}\ge 0$

Question

Let $a,b,c$ are nonnegative real numbers such that $\frac1{a^3+1}+\frac1{b^3+1}+\frac1{c^3+1}+\frac1{d^3+1}=2$.

Prove the inequality $$\frac{1-a}{a^2-a+1}+\frac{1-b}{b^2-b+1}+\frac{1-c}{c^2-c+1}+\frac{1-d}{d^2-d+1}\ge0$$

My attempts:

Let $$A=\frac{1-a}{a^2-a+1}+\frac{1-b}{b^2-b+1}+\frac{1-c}{c^2-c+1}+\frac{1-d}{d^2-d+1}$$

If $a=1$ then $A=0$ then inequality is hold. If $a\not=1$ then

$$A=\frac{1-a^2}{a^3+1}+\frac{1-b^2}{b^3+1}+\frac{1-c^2}{c^3+1}+\frac{1-d^2}{d^3+1}$$

I tried the method of Lagrange multipliers and Jensen's inequality but I have not been proved this inequality

Answer 1

With the condition $\sum\limits_{cyc}\frac{1}{a^3+1}=2$ it's true: $$\sum_{cyc}\frac{1-a}{1-a+a^2}=\sum_{cyc}\left(\frac{1-a}{1-a+a^2}+\frac{4}{3}\left(\frac{1}{2}-\frac{1}{a^3+1}\right)\right)=\sum_{cyc}\frac{(a-1)^2(2a+1)}{3(a^3+1)}\geq0.$$

Answer 2

Since $$ \sum_{k=1}^4\frac1{1+a_k^3}=2 $$ we have $$ \sum_{k=1}^4\frac{a_k^3}{1+a_k^3}=2 $$ By Holder's Inequality, with $p=\frac32$ and $q=3$, $$ \begin{align} \sum_{k=1}^4\frac{a_k^2\cdot1}{1+a_k^3} &\le\left(\sum_{k=1}^4\frac{a_k^3}{1+a_k^3}\right)^{2/3}\left(\sum_{k=1}^4\frac1{1+a_k^3}\right)^{1/3}\\ &=2 \end{align} $$ Therefore, $$ \begin{align} \sum_{k=1}^4\frac{1-a_k}{1-a_k+a_k^2} &=\sum_{k=1}^4\frac{1-a_k^2}{1+a_k^3}\\ &\ge2-2\\[9pt] &=0 \end{align} $$