<s>[INST] $ x+y+z = 3, \; \sum\limits_{cyc} \frac{x}{2x^2+x+1} \leq \frac{3}{4} $ 
For positive variables $ x+y+z=3 $, show that $ \displaystyle \sum_{cyc} \dfrac{x}{2x^2+x+1} \leq \dfrac{3}{4} $.

Apart from $ (n-1)$ EV, I could not prove this inequality. I've tried transforming it into a more generic problem - it looks fairly more interesting to me.

Consider a continuous function $ f(x) \geq 0, \; 0 < x < \alpha $, with a unique value of $ \gamma $ such that $ f'(\gamma) = 0 $ and a unique value of $ \theta $ such that $ f''(\theta) = 0 $, being $ \gamma < \theta $.
If $ \displaystyle \sum_{i=1}^{n} x_i = \alpha, \; f(0)=0$ and $ f \left ( \frac{\alpha}{n} \right ) = \frac{\beta}{n} $, is it always correct to conclude that $ \displaystyle \sum_{i=1}^{n} f (x_i) \leq \beta $, being $ f(\gamma) > \frac{\beta}{n} $? Provide an example other than $ f(x) = \frac{x}{2x^2+x+1} $ and a counter-example.

 [/INST] Also, the Tangent Line method helps.
Indeed, we need to prove that:
$$\sum_{cyc}\left(\frac{1}{4}-\frac{x}{2x^2+x+1}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)(2x-1)}{2x^2+x+1}\geq0$$ or
$$\sum_{cyc}\left(\frac{(x-1)(2x-1)}{2x^2+x+1}-\frac{x-1}{4}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)^2(5-2x)}{2x^2+x+1}\geq0,$$
which says that our inequality is true for $\max\{x,y,z\}\leq\frac{5}{2}.$
Now, let $x>\frac{5}{2}.$
Thus, since by AM-GM $$\frac{y}{2y^2+y+1}=\frac{1}{2y+\frac{1}{y}+1}\leq\frac{1}{2\sqrt2+1},$$ we obtain:
$$\sum_{cyc}\frac{x}{2x^2+x+1}\leq\frac{2}{2\sqrt2+1}+\frac{2.5}{2\cdot2.5^2+2.5+1}<\frac{3}{4}.$$
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