Conjecture for the inequality $f\left(x\right)+f\left(y\right)+f\left(z\right)\ge1$ where $f\left(x\right)=\frac{x}{x+\frac{1}{x}+1}$ which seems easy

Question

I come back with an inequality checked with Desmos :

Let $x\in R^*$ then define :

$$f\left(x\right)=\frac{x}{x+\frac{1}{x}+1}$$

Then do we have :

$$f\left(x\right)+f\left(y\right)+f\left(z\right)\ge1$$

Where $xyz=1$

I have tried to use the substitution $x=e^y$ but the resulting function $f(e^y)$ is neither convex or concave on $(-\infty,\infty)$

On the other hand it seems easier to use the negative value as for example $f(x)\geq 1$ for $x\leq -1$.

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Answer using a computer (Desmos +Geogebra) :

$$a(x,y)=f\left(x^{2}\right)+f\left(y^{2}\right)-2f\left(xy\right)\geq^?0\tag{I}$$

Then :

$$a(x/y,y/x*z/(z+1))=(x^{2}z+x^{2}-zy^{2})^{2}(x^{4}z^{4}+5x^{4}z^{3}+8x^{4}z^{2}+5x^{4}z+x^{4}+x^{2}z^{4}y^{2}+7x^{2}z^{3}y^{2}+7x^{2}z^{2}y^{2}+2x^{2}zy^{2}+z^{4}y^{4}+3z^{3}y^{4}+z^{2}y^{4})/((3z^{2}+3z+1)(x^{2}-xy+y^{2})(x^{2}+xy+y^{2})(x^{2}z^{2}+2x^{2}z+x^{2}-xz^{2}y-xzy+z^{2}y^{2})(x^{2}z^{2}+2x^{2}z+x^{2}+xz^{2}y+xzy+z^{2}y^{2}))$$

Then we conclude that ${I}$ is true for $0<xy\leq 1$ and $x,y\in(0,\infty)$

Remains to show it as a one variable inequality .

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How to prove nicely this inequality ?

Can we find a more difficult inequality (constraint or coefficient)?

Thanks .

Answer 1

Another way.

Let $x=\frac{a}{b}$ and $y=\frac{b}{c}$, where $a$, $b$ and $c$ are positives.

Thus, $z=\frac{c}{a}$ and by C-S and Rearrangement we obtain: $$\sum_{cyc}\frac{x}{x+\frac{1}{x}+1}=\sum_{cyc}\frac{a^2}{a^2+ab+b^2}=\sum_{cyc}\frac{a^2(a+c)^2}{(a^2+ab+b^2)(a+c)^2}\geq$$ $$\geq\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}(a^2+ab+b^2)(a+c)^2}=\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}(a^4+a^3b+2a^3c+3a^2b^2+5a^2bc)}\geq$$ $$\geq\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+4a^2bc)}=1.$$ We have proved that for any positives $a$, $b$ and $c$ the following inequality is true. $$\sum\limits_{cyc}\frac{a^2}{a^2+ab+b^2}\geq1.$$

The following inequality is stronger.

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\tfrac{a^2}{a^2+ab+b^2}+\tfrac{b^2}{b^2+bc+c^2}+\tfrac{c^2}{c^2+ca+a^2}\geq1+\tfrac{20abc(a-b)^2(a-c)^2(b-c)^2}{(a+b+c)^3(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)}.$$

Answer 2

Because by Muirhead $$\sum_{cyc}\frac{x}{x+\frac{1}{x}+1}-1=\frac{\sum\limits_{cyc}(x^2y^2-x)}{\prod\limits_{cyc}(x^2+x+1)}=\frac{\sum\limits_{cyc}(x^2y^2-x^2yz)}{\prod\limits_{cyc}(x^2+x+1)}\geq0.$$