$\frac34\le\sum_\text{cyc}\big(\frac{x}{x+1}\big)^2<2\ $ for $x,y,z>0$ and $xyz=1$.

Question

For $x,y,z>0$ and $xyz=1$, what is the range of $f(x,y,z)$? $$f(x,y,z):=\Big(\frac{x}{x+1}\Big)^2+\Big(\frac{y}{y+1}\Big)^2+\Big(\frac{z}{z+1}\Big)^2.$$

I conjecture $\min f=\frac34$ at $(x,y,z)=(1,1,1)$ and $\sup f=2$ at $x\to\infty,y\to\infty, z=\frac1{xy}$. How would one prove this?

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I tried using the substitution $x=\frac ab, y=\frac bc, z=\frac ca$, for some positive $a, b, c$, but did not see a clear route ahead.

Answer 1

We'll prove that $$\sum_{cyc}\frac{x^2}{(x+1)^2}\geq\frac{3}{4}.$$ Indeed, let $x=\frac{a}{b}$ and $y=\frac{b}{c},$ where $a$, $b$ and $c$ are positives.

Thus, $z=\frac{c}{a}$ and we need to prove that: $$\sum_{cyc}\frac{a^2}{(a+b)^2}\geq\frac{3}{4}.$$ Now, by C-S $$\sum_{cyc}\frac{a^2}{(a+b)^2}=\sum_{cyc}\frac{a^2(a+c)^2}{(a+b)^2(a+c)^2}\geq\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}(a+b)^2(a+c)^2}$$ and it's enough to prove that: $$4\left(\sum\limits_{cyc}(a^2+ab)\right)^2\geq3\sum\limits_{cyc}(a+b)^2(a+c)^2,$$ which is fourth degree and it's true by $uvw$ even for any real variables:

Let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v^2$ may be negative, and $abc=w^3$.

Thus, we need to prove a linear inequality of $w^3,$ which says that it's enough to prove the last inequality for an extremal value of $w^3$, which happens for equality case of two variables.

If $b=c=0$, the inequality is obvious.

If $b=c=1$ we obtain $$(a-1)^2(a+3)^2\geq0,$$ which ends a proof of this inequality.

About $uvw$ see here: https://artofproblemsolving.com/community/c6h278791

Also, $$\sum_{cyc}\frac{a^2}{(a+b)^2}<\frac{a^2}{a^2+b^2}<\sum_{cyc}\frac{a^2+c^2}{a^2+b^2+c^2}=2$$ and by your work we got an infimum and a supremum of $f$,

which by continuity gives a range: $\left[\frac{3}{4},2\right).$

Answer 2

Remark: pqr method for $\frac34 \le f < 2$:

Let $p = x + y + z, q = xy + yz + zx, r = xyz = 1$.

To find the pqr expression of $f$, let $A = \frac{x}{x + 1} + \frac{y}{y + 1} + \frac{z}{z + 1}$ and $B = \frac{x}{x + 1}\frac{y}{y + 1} + \frac{y}{y + 1}\frac{z}{z + 1} + \frac{z}{z + 1}\frac{x}{x + 1}$. We have $$A = \frac{3xyz + 2(xy + yz + zx) + (x + y + z)}{(x + 1)(y + 1)(z + 1)} = \frac{3r + 2q + p}{r + q + p + 1} = \frac{3 + 2q + p}{p + q + 2},$$ and $$B = \frac{3xyz + (xy + yz + zx)}{(x+1)(y+1)(z+1)} = \frac{3r + q}{r + q + p + 1} = \frac{3 + q}{p + q + 2}.$$ Then we have $$f = A^2 - 2B = \frac{p^2 + 2pq + 2q^2 + 2q - 3}{(p+q+2)^2}.$$

(1) $f \ge \frac34$ is equivalent to $$5q^2 + (2p-4)q + p^2 - 12p - 24 \ge 0. \tag{1}$$

Using $p^3 \ge 27r$, we have $p \ge 3$. Using $q^2 \ge 3pr$, it suffices to prove that $$5\cdot 3p + (2p-4)\sqrt{3p} + p^2 - 12p - 24 \ge 0$$ or $$(p+12)(p - 3) + (2p - 4)(\sqrt{3p} - 3) \ge 0$$ which is true.

(2) We have $$2 - f = \frac{p^2 + 2pq + 8p + 6q + 11}{(p + q + 2)^2} > 0.$$

We are done.