$ x+y+z = 3, \; \sum\limits_{cyc} \frac{x}{2x^2+x+1} \leq \frac{3}{4} $

Question

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.

Answer 1

$\left(\frac{x}{2x^2+x+1}\right)''=\frac{2(4x^3-6x-1)}{(2x^2+x+1)^3}<0$ 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!

Answer 2

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: $$\sum_{cyc}\frac{x}{2x^2+x+1}\leq\frac{2}{2\sqrt2+1}+\frac{2.5}{2\cdot2.5^2+2.5+1}<\frac{3}{4}.$$

Answer 3

Another way.

We need to prove that: $$\frac{3}{4}-\sum_{cyc}\frac{3x(x+y+z)}{18x^2+3x(x+y+z)+(x+y+z)^2}\geq0$$ or $$\sum_{cyc}\left(\frac{1}{4}-\frac{3x(x+y+z)}{18x^2+3x(x+y+z)+(x+y+z)^2}\right)\geq0$$ or $$\sum_{cyc}\frac{10x^2-7xy-7xz+y^2+2yz+z^2}{2x^2+x+1}\geq0$$ or $$\sum_{cyc}\frac{(x-y)(5x-y-z)-(z-x)(5x-y-z)}{2x^2+x+1}\geq0$$ or $$\sum_{cyc}(x-y)\left(\frac{5x-y-z}{2x^2+x+1}-\frac{5y-x-z}{2y^2+y+1}\right)\geq0$$ or $$\sum_{cyc}(x-y)^2(2z^2+z+1)(2x^2-8xy+2y^2+2xz+2yz+x+y+z+6)\geq0$$ or $$\sum_{cyc}(x-y)^2(2z^2+z+1)(3x^2-6xy+3y^2+z^2+4xz+4yz)\geq0,$$ which is obvious.