Inequality $\sum_{cyc}\frac{a}{2a^2+a+1}\leq \frac{3}{4}$

Question

Let $a,b,c\in\mathbb{R^+}$ such that $a+b+c=3$. Then prove that $$\sum_{cyc}\frac{a}{2a^2+a+1}\leq \frac{3}{4}$$

I tried to use tangent line method. Let $$f(x)=\frac{x}{2x^2+x+1}$$ Then $$f'(x)=\frac{1-2x^2}{2x^2+x+1}$$ and since we know that equality occurs if $a=b=c=1$, tangent line is: $$y=f(1)+(x-1)f'(1)=\frac{5-x}{16}$$ But $$\frac{x}{2x^2+x+1}\leq\frac{5-x}{16}\implies x\in(-\infty,\frac{5}{2}]$$ it means this is false for $x\in(\frac{5}{2},3]$. Then how to prove original inequality for $max(a,b,c)\geq\frac{5}{2}$? Can anyone help me?

Answer 1

This inequality is true for any reals $a$, $b$ and $c$ such that $a+b+c=3$.

Indeed, if $abc=0$ so the inequality is obvious.

But for $abc\neq0$, by AM-GM and your work (after assuming that $a\geq\frac{5}{2}$ and $x\neq0$ ) we obtain: $$\sum_{cyc}\frac{a}{2a^2+a+1}\leq\frac{\frac{5}{2}}{2\left(\frac{5}{2}\right)^2+\frac{5}{2}+1}+\frac{2|x|}{2x^2+|x|+1}=$$ $$=\frac{5}{32}+\frac{2}{2|x|+\frac{1}{|x|}+1}\leq\frac{5}{32}+\frac{2}{2\sqrt2+1}<\frac{3}{4}.$$