$ 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.


$\left(\frac{x}{2x^2+x+1}\right)''=\frac{2(4x^3-6x-1)}{(2x^2+x+1)^3}<0$ for all $0<x<1$.
Hence, by Vasc's LCF Theorem it remains to prove our inequality for $y=x$ and $z=3-2x$,
which gives $(x-1)^2(8x^2-14x+9)\geq0$. Done!
