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Let $x,y,z>0$. Prove that: $$3\geq \frac{(x+ y)^{2}x^{2}}{(x^{2}+ y^{2})^{2}}+ \frac{(y+ z)^{2}y^{2}}{(y^{2}+ z^{2})^{2}}+ \frac{(z+ x)^{2}z^{2}}{(z^{2}+ x^{2})^{2}}$$

I need to the hints and hope to see the Buffalo Way help here! Thanks a lot!

My idea is as follows: Because this inequality is cyclic. So, it's enough to prove this inequality in two cases: $x\leq y\leq z$ and $x\geq y\geq z$. I can prove it with $x\leq y\leq z$ but with $x\geq y\geq z$, I can't.

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    $\begingroup$ Proving any non-trivial inequality with the Buffalo way is not fun. $\endgroup$ Jan 29, 2020 at 1:23
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    $\begingroup$ Since the RHS is homogeneous, we may let either $x^2+y^2+z^2=1$ or $x+y+z=1,$ whichever helps us the most. $\endgroup$ Jan 29, 2020 at 1:24
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    $\begingroup$ Another idea: Let $u=y/x, v=z/y, w=x/z$ so that the inequality becomes $\sum\limits_{\text{cyc}} \frac{(1+u)^2}{(u^2+1)^2} \le 3$ subject to $uvw=1.$ $\endgroup$ Jan 29, 2020 at 1:26
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    $\begingroup$ My idea is as follows: Because this inequality is cyclic. So, it's enough to prove this inequality in two cases: $x \le y \le z$ and $x\geq y \geq z$. I can prove it with $x \le y\leq z$ but with $x\geq y\geq z$, I can't. $\endgroup$
    – NKellira
    Jan 29, 2020 at 1:33
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    $\begingroup$ @Display name BW does not help here! The Vasc's theorems don't work! $\endgroup$ Jan 29, 2020 at 5:20

3 Answers 3

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Remark: Let us describe simply a known trick for the problem of proving $f(u)+f(v)+f(w)\ge 0$ under constraint $uvw=1$ and $u, v, w > 0$.

The method of Lagrange multipliers yields the system of equations \begin{align} f'(u) &= \lambda vw, \\ f'(v) &= \lambda uw, \\ f'(w) &= \lambda uv,\\ uvw &= 1.\tag{1} \end{align} Clearly, we have $uf'(u) = vf'(v) = wf'(w) = \lambda$. If the equation $xf'(x) = c$ has at most two distinct positive real solutions for any $c \in \mathbb{R}$, then two of $u, v, w$ are equal.

This trick is useful for many problems. For example,

Example 1: Let $a, b, c$ be positive real numbers such that $abc=1$. Prove that $$\frac{a}{a^{11}+1} + \frac{b}{b^{11} + 1} + \frac{c}{c^{11}+1} \le \frac{3}{2}.$$

Example 2: Let $a, b, c$ be positive real numbers such that $abc=1$. Prove that $$\frac{7-6a}{2+a^2} + \frac{7-2b}{2+b^2} + \frac{7-2c}{2+c^2} \ge 1.$$

Example 3: Let $a, b, c$ be positive real numbers such that $abc=1$. Prove that $$\frac{1}{a+3} + \frac{1}{b+3} + \frac{1}{c+3} \ge \frac{a}{a^2+3} + \frac{b}{b^2+3} + \frac{c}{c^2+3}.$$

Example 4: Let $a, b, c$ be positive real numbers such that $abc=1$. Prove that $$\frac{1}{a+4} + \frac{1}{b+4} + \frac{1}{c+4} \ge \frac{a}{a^2+4} + \frac{b}{b^2+4} + \frac{c}{c^2+4}.$$

Example 5: Let $a, b, c$ be positive real numbers such that $abc=1$. Prove that $$\sum_{\mathrm{cyc}} \sqrt{\frac{a}{a+8}} \ge 1.$$

$\phantom{2}$

Use this trick for the OP:

Equivalent problem (as @Display name pointed out): Let $u, v, w > 0$ with $uvw=1$. Prove that $$\frac{(1+u)^2}{(1+u^2)^2}+\frac{(1+v)^2}{(1+v^2)^2}+\frac{(1+w)^2}{(1+w^2)^2}\leq 3.$$

Let $f(x) = 1 - \frac{(1+x)^2}{(1+x^2)^2}$. We need to prove that $f(u)+f(v)+f(w)\ge 0$. The method of Lagrange multipliers yields the system of equations \begin{align} f'(u) &= \lambda vw, \\ f'(v) &= \lambda uw, \\ f'(w) &= \lambda uv, \\ uvw &= 1. \end{align} Clearly, we have $uf'(u) = vf'(v) = wf'(w) = \lambda$. Let us prove that if $(u, v, w, \lambda)$ with $u, v, w > 0$ satisfies the above system of equations, then two of $u, v, w$ are equal.

Let $F(x) = xf'(x) = \frac{2x(1+x)(x^2+2x-1)}{(x^2+1)^3}$.

Clearly, $F(0) = 0$, $F(\sqrt{2}-1) = 0$, $F(x) < 0$ on $(0, \sqrt{2}-1)$, and $F(x) > 0$ on $(\sqrt{2}-1, +\infty)$.

We have $F'(x) = -\frac{2(2x^5+9x^4-14x^2-2x+1)}{(x^2+1)^4}$. Let $G(x) = 2x^5+9x^4-14x^2-2x+1$. From Descartes' sign rule, since there are two sign changes, $G(x) = 0$ has at most two positive real roots. Also, we have $G(0) > 0$, $G(\sqrt{2}-1) = 32-24\sqrt{2} < 0$ and $G(+\infty) = +\infty$. Thus, $G(x) = 0$ has exactly one real solution on $(0, \sqrt{2}-1)$ and $(\sqrt{2}-1, +\infty)$, respectively. Thus, $F'(x) = 0$ has exactly one real solution on $(0, \sqrt{2}-1)$ and $(\sqrt{2}-1, +\infty)$, respectively.

Figure of $F(x)$: Figure of F

Thus, $F(x) = c$ has at most two distinct positive real solutions for any real number $c$. Since $F(u) = F(v) = F(w)$, we know that two of $u, v, w$ are equal.

For $u = v > 0$ and $w = \frac{1}{u^2}$, it is easy to prove that \begin{align} &f(u) + f(u) + f(\frac{1}{u^2})\\ =\ & \frac{(2u^{10}+4u^9+6u^8+4u^7+3u^6+2u^5+5u^4-u^2-2u+1)(u-1)^2}{(u^2+1)^2(u^4+1)^2}\\ \ge & 0. \end{align} Thus, the inequality is true for $u, v, w > 0$ satisfying the system of equations (1).

It remains to prove that the inequality is true if $\min(u, v, w) \to 0^{+}$ (meaning $(u,v,w)$ approaches the boundary of the constraint).

Let $H(x) = \frac{(1+x)^2}{(1+x^2)^2}$. It is easy to prove that $H(x) \le \frac{147}{100}$ for all real numbers $x$. Note also that $H(x) \le \frac{3}{100}$ for $x \ge 10$. Thus, if $\min(u, v, w) \to 0^{+}$, then $H(u) + H(v) + H(w) \le \frac{147}{100} + \frac{147}{100} + \frac{3}{100} = \frac{297}{100}$. The desired result follows.

We are done.

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    $\begingroup$ Starting from $F(x)=F(y)=F(z)$ (and from $xyz=1$) why do we have the equality of two of the three variables $x,y,z$? Note that the factorization of $F(x)-F(y)$ gives a numerator of the shape $$(x^5 y^4 + x^4 y^5 + 3 \, x^5 y^3 + 3 \, x^4 y^4 + 3 \, x^3 y^5 + x^5 y^2 + x^4 y^3 + x^3 y^4 + x^2 y^5 - x^5 y - x^4 y^2 + 8 \, x^3 y^3 - x^2 y^4 - x y^5 - 3 \, x^3 y - 12 \, x^2 y^2 - 3 \, x y^3 - x^3 - x^2 y - x y^2 - y^3 - 3 \, x^2 - 6 \, x y - 3 \, y^2 - x - y + 1)\\ \ \cdot\ (x - y)$$ and it is unclear why the first factor is $\ne 0$. $\endgroup$
    – dan_fulea
    Feb 10, 2020 at 12:10
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    $\begingroup$ @dan_fulea Because $F(u) = c$ has at most two positive real roots for any real number $c$. If $x, y, z$ are distinct and $F(x) = F(y) = F(z) $ ($=c$), then $F(u)=c$ has three distinct positive real roots, contradiction. $\endgroup$
    – River Li
    Feb 10, 2020 at 12:21
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    $\begingroup$ Yes, good work, 1+ $\endgroup$
    – dan_fulea
    Feb 11, 2020 at 10:41
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    $\begingroup$ Do you have any stronger form for this inequality? $\endgroup$
    – NKellira
    Feb 26, 2021 at 14:21
  • $\begingroup$ @tthnew For $3\geq \frac{(x+ y)^{2}x^{2}}{(x^{2}+ y^{2})^{2}}+ \frac{(y+ z)^{2}y^{2}}{(y^{2}+ z^{2})^{2}}+ \frac{(z+ x)^{2}z^{2}}{(z^{2}+ x^{2})^{2}}$? I don't have simple stronger one. But I think you can get some form $3\geq \frac{(x+ y)^{2}x^{2}}{(x^{2}+ y^{2})^{2}}+ \frac{(y+ z)^{2}y^{2}}{(y^{2}+ z^{2})^{2}}+ \frac{(z+ x)^{2}z^{2}}{(z^{2}+ x^{2})^{2}} + C\frac{(x-y)^2(y-z)^2(z-x)^2}{(x^2+y^2+z^2)^3}$ for some constant $C$. $\endgroup$
    – River Li
    Feb 26, 2021 at 15:10
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Partial answer

Following an idea of Display name we have to show :

Let $u,v,w>0$ such that $uvw=1$ then we have : $$\frac{(1+u)^2}{(1+u^2)^2}+\frac{(1+v)^2}{(1+v^2)^2}+\frac{(1+w)^2}{(1+w^2)^2}\leq 3$$

The main idea is to use trigonometry :

Let $u=\tan(\frac{x}{2})$ and $v=\tan(\frac{y}{2})$ and $w=\tan(\frac{z}{2})$

The inequality becomes :

$$\frac{(1+\tan(\frac{x}{2}))^2}{(1+\tan^2(\frac{x}{2}))^2}+\frac{(1+\tan(\frac{y}{2}))^2}{(1+\tan^2(\frac{y}{2}))^2}+\frac{(1+\tan(\frac{z}{2}))^2}{(1+\tan^2(\frac{w}{2}))^2}\leq 3$$

But we have the following relation putting $t=\tan(\frac{x}{2})$ (Weierstrass substitution):

$$\sin(x)=\frac{2t}{1+t^2}$$

$$\cos(x)+1=\frac{2}{1+t^2}$$

So we have :

$$\frac{\cos(x)+\sin(x)+1}{2}=\frac{1+t}{1+t^2}$$

Putting this in the inequality we have to show :

$$\Big(\frac{\cos(x)+\sin(x)+1}{2}\Big)^2+\Big(\frac{\cos(y)+\sin(y)+1}{2}\Big)^2+\Big(\frac{\cos(z)+\sin(z)+1}{2}\Big)^2\leq 3$$

We study the second derivative of the function :

$$f(x)=\Big(\frac{\cos(x)+\sin(x)+1}{2}\Big)^2$$

Which is equal to :

$$f''(x)=-\frac{\sin(x)}{2} - \frac{\cos(x)}{2} - 2 \sin(x) \cos(x)$$ The function $f(x)$ is concave on $[0,p]$ where $p$ have the value :

$$p = 2 \Big(- \tan^{-1}\Big(\frac{3}{2} - \frac{\sqrt{17}}{2} - \sqrt{0.5 (5 - \sqrt{17})}\Big)\Big)>\frac{\pi}{2}$$

So we can apply Jensen's inequality for $x,y,z\in [0,p]$ we have :

$$\sum_{cyc}\Big(\frac{\cos(x)+\sin(x)+1}{2}\Big)^2\leq 3\Big(\frac{\cos(\frac{x+y+z}{3})+\sin(\frac{x+y+z}{3})+1}{2}\Big)^2$$

Or :

$$\sum_{cyc}\Big(\frac{\cos(x)+\sin(x)+1}{2}\Big)^2\leq 3\frac{(1+\tan(\frac{x+y+z}{6}))^2}{(1+\tan^2(\frac{x+y+z}{6}))^2}$$

With the conditions : $0<x<\pi$ and $0<y<\pi$ and $0<z<\pi$ and $\tan(\frac{x}{2})\tan(\frac{y}{2})\tan(\frac{z}{2})=1$

Second edit :

As pointed out by River Li I add a restriction we need to have :

$\frac{3\pi}{4}\leq \frac{x+y+z}{2}$ with the condition $\tan(\frac{x}{2})\tan(\frac{y}{2})\tan(\frac{z}{2})=1$

Or :

$$\frac{3\pi}{4}\leq\tan^{-1}(a)+\tan^{-1}(b)+\tan^{-1}(c) $$

With $a=\tan(\frac{x}{2})$ and $b=\tan(\frac{y}{2})$ and $c=\tan(\frac{z}{2})$

Now if $\max(a,b,c)=a$ and $\min(a,b,c)=c$ we add the restriction $ab>1$ and we have $\frac{a+b}{1-ab}c<0<1$

So we have :

$$\tan^{-1}(a)+\tan^{-1}(b)=\tan^{-1}\Big(\frac{a+b}{1-ab}\Big)+\pi$$

And :

$$\tan^{-1}(a)+\tan^{-1}(b)+\tan^{-1}(c)=\tan^{-1}\Big(\frac{a+b+c-abc}{1-ab-bc-ca}\Big)+\pi$$

So we need to have :

$$\frac{3\pi}{4}\leq\tan^{-1}\Big(\frac{a+b+c-abc}{1-ab-bc-ca}\Big)+\pi$$

Or :

$$\frac{-\pi}{4}\leq\tan^{-1}\Big(\frac{a+b+c-1}{1-ab-bc-ca}\Big)$$

Or :

$$-1\leq \frac{a+b+c-1}{1-ab-bc-ca}$$

Or :

$$1\geq \frac{1-(a+b+c)}{1-ab-bc-ca}$$ Or :

$$1-(a+b+c)\geq 1-ab-bc-ca$$

Or

$$(a+b+c)\leq ab+bc+ca$$

End of the second edit .

So we have :$\frac{3\pi}{2}\leq x+y+z\leq2\pi$ or $\frac{3\pi}{12}\leq\frac{x+y+z}{6}\leq\frac{\pi}{3}$

So $1\leq\tan(\frac{x+y+z}{6})$

But the function $g(x)=\frac{(1+x)^2}{(1+x^2)^2}$ is decreasing on $[1,\infty]$

So $$\sum_{cyc}\Big(\frac{\cos(x)+\sin(x)+1}{2}\Big)^2\leq3\frac{(1+\tan(\frac{x+y+z}{6}))^2}{(1+\tan^2(\frac{x+y+z}{6}))^2}\leq 3$$

And we are done .

Edit:

Since $$(\frac{\sin(x)+\cos(x)+1}{2})^2=0.5+\frac{\sin(x)+\cos(x)+\sin(x)\cos(x)}{2}$$

We have to show with the conditions : $0<x<\pi$ and $0<y<\pi$ and $0<z<\pi$ and $\tan(\frac{x}{2})\tan(\frac{y}{2})\tan(\frac{z}{2})=1$

:

$$\sum_{cyc}\frac{\sin(x)+\cos(x)+\sin(x)\cos(x)}{2}\leq 1.5$$

But since $$\sum_{cyc}\frac{\sin(x)+\cos(x)}{2}\leq 1.5$$

Because $h(x)=\sin(x)+\cos(x)$ is concave on $[0,0.75\pi]$ we have:

$$\sum_{cyc}\frac{\sin(x)+\cos(x)}{2}\leq 3\frac{\sin(\frac{x+y+z}{3})+\cos(\frac{x+y+z}{3})}{2}$$

And the same reasoning as below conducts to

$$\sum_{cyc}\frac{\sin(x)+\cos(x)}{2}\leq 3\frac{\sin(\frac{x+y+z}{3})+\cos(\frac{x+y+z}{3})}{2}\leq 1.5$$

Remains to show that :

$$(\sin(2x)+\sin(2y)+\sin(2z))0.25\leq 0$$

Can you end now ?

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    $\begingroup$ Would you please check: $x = \frac{14685}{10000}, \ y = \frac{17700}{10000}$ and $z = 2\arctan\frac{1}{\tan(\frac{1}{2}x) \tan (\frac{1}{2}y)} \approx 1.472893942$, then $\tan\frac{x}{2}\tan\frac{y}{2}\tan\frac{z}{2}=1$, $x, y, z \in (0, p)$ where $p \approx 1.771322936$, however, $\tan\frac{x+y+z}{6} \approx 0.9996683754 < 1$. Am I missing something? $\endgroup$
    – River Li
    Jan 31, 2020 at 12:58
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    $\begingroup$ @RiverLi yes there is a problem .It's not total wrong because there are examples where it's works .So we have to add a condition .Thanks for the remark . $\endgroup$ Jan 31, 2020 at 15:18
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    $\begingroup$ @RiverLi Please check my second edit .Thanks again. $\endgroup$ Jan 31, 2020 at 16:14
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    $\begingroup$ So, you prove the case when $\tan\frac{x}{2}\tan\frac{y}{2}\tan\frac{z}{2}=1$, $x, y, z \in (0, p)$ and $x+y+z \ge \frac{3\pi}{2}$, right? If so, these conditions leads to $abc=1$, $a, b,c \in (0.67, 1.22)$ (approximately, I ran numerical simulation to find the values), right? $\endgroup$
    – River Li
    Feb 1, 2020 at 1:31
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    $\begingroup$ @RiverLi yes with the equivalent restriction $a+b+c\leq ab+bc+ca$ .I work now to get the maximum of the function $g(x)=\frac{(1+x)^2}{(1+x^2)^2}$ in my reasoning. $\endgroup$ Feb 1, 2020 at 14:20
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It's possible to prove by Buffalo Way, at least by helping of CAS.

Let $f(x,y,z)=\sum\limits_\text{cyc}{\dots}-3$, then it's easy to prove $x\le y\le z$ by $y=1+u$ and $z=1+u+v$.

The case $x\le z\le y$ is more complicated, because we get negative terms:

$$g=4u^{12}+\dots-10u^5v^5-\dots+32v^2$$

which need to apply BW again by $v=u(1+v')$ for $u\le v$ and $u=v(1+u')$ for $v\le u$.

In the case $u\le v$, we still get negative terms:

$$h=u^{10}v^8+\dots-844u^7v^4-\dots+96$$

This time we can't use the same trick because there is 0-degree term in the polynomial.

Then we need to take $h$ as piecewise: set $u=\dfrac{1}{1+u'}$ for $u\le 1$, $u=1+u'$ for $u\ge 1$, $v=\dfrac{1}{1+v'}$ for $v\le 1$ and $v=1+v'$ for $v\ge 1$.

Finally, we get all positive terms for $h(u\le 1,v\le 1)$, $h(u\le 1,v\ge 1)$, $h(u\ge 1,v\le 1)$ and $h(u\ge 1,v\ge 1)$.

Here is my code by SymPy:

from sympy import *

def cyc(f, vars):
    x, y, z = vars
    t = symbols('t', positive = True)
    return f.subs(z, t).subs(y, z).subs(x, y).subs(t, x)

def sum_cyc(f, vars):
    f1 = cyc(f, vars)
    return f + f1 + cyc(f1, vars)

def main():
    x, y, z = symbols('x, y, z', positive = True)
    f = sum_cyc((x + y)**2*x**2/(x**2 + y**2)**2, (x, y, z)) - 3
    u, v = symbols('u, v', positive = True)
    print('f(xyz) =', factor(f.subs(y, x*(1 + u)).subs(z, x*(1 + u + v))))
    print('f(xzy) =', factor(f.subs(z, x*(1 + u)).subs(y, x*(1 + u + v))))
    # result in f(xzy)
    g = 4*u**12 + 20*u**11*v + 40*u**11 + 43*u**10*v**2 + 172*u**10*v + 192*u**10 + 54*u**9*v**3 + 302*u**9*v**2 + 692*u**9*v + 576*u**9 + 47*u**8*v**4 + 286*u**8*v**3 + 935*u**8*v**2 + 1700*u**8*v + 1188*u**8 + 32*u**7*v**5 + 174*u**7*v**4 + 548*u**7*v**3 + 1624*u**7*v**2 + 2816*u**7*v + 1752*u**7 + 17*u**6*v**6 + 88*u**6*v**5 + 96*u**6*v**4 + 228*u**6*v**3 + 1688*u**6*v**2 + 3296*u**6*v + 1864*u**6 + 6*u**5*v**7 + 42*u**5*v**6 - 10*u**5*v**5 - 524*u**5*v**4 - 728*u**5*v**3 + 1088*u**5*v**2 + 2792*u**5*v + 1408*u**5 + u**4*v**8 + 14*u**4*v**7 + 20*u**4*v**6 - 276*u**4*v**5 - 1080*u**4*v**4 - 1152*u**4*v**3 + 588*u**4*v**2 + 1728*u**4*v + 720*u**4 + 2*u**3*v**8 + 16*u**3*v**7 - 4*u**3*v**6 - 248*u**3*v**5 - 640*u**3*v**4 - 416*u**3*v**3 + 512*u**3*v**2 + 768*u**3*v + 224*u**3 + 3*u**2*v**8 + 24*u**2*v**7 + 84*u**2*v**6 + 160*u**2*v**5 + 252*u**2*v**4 + 384*u**2*v**3 + 432*u**2*v**2 + 224*u**2*v + 32*u**2 + 4*u*v**8 + 40*u*v**7 + 152*u*v**6 + 328*u*v**5 + 448*u*v**4 + 384*u*v**3 + 192*u*v**2 + 32*u*v + 4*v**8 + 24*v**7 + 72*v**6 + 128*v**5 + 144*v**4 + 96*v**3 + 32*v**2
    print('g(uv) =', factor(g.subs(v, u*(1 + v))))
    print('g(vu) =', factor(g.subs(u, v*(1 + u))))
    # result in g(uv)
    h = u**10*v**8 + 14*u**10*v**7 + 87*u**10*v**6 + 316*u**10*v**5 + 742*u**10*v**4 + 1168*u**10*v**3 + 1216*u**10*v**2 + 768*u**10*v + 224*u**10 + 2*u**9*v**8 + 30*u**9*v**7 + 196*u**9*v**6 + 746*u**9*v**5 + 1874*u**9*v**4 + 3304*u**9*v**3 + 4064*u**9*v**2 + 3136*u**9*v + 1120*u**9 + 3*u**8*v**8 + 40*u**8*v**7 + 216*u**8*v**6 + 614*u**8*v**5 + 1116*u**8*v**4 + 1960*u**8*v**3 + 3775*u**8*v**2 + 4796*u**8*v + 2492*u**8 + 4*u**7*v**8 + 56*u**7*v**7 + 276*u**7*v**6 + 428*u**7*v**5 - 844*u**7*v**4 - 3644*u**7*v**3 - 3040*u**7*v**2 + 2332*u**7*v + 3352*u**7 + 4*u**6*v**8 + 72*u**6*v**7 + 476*u**6*v**6 + 1320*u**6*v**5 + 620*u**6*v**4 - 4224*u**6*v**3 - 7244*u**6*v**2 - 736*u**6*v + 3764*u**6 + 24*u**5*v**7 + 320*u**5*v**6 + 1576*u**5*v**5 + 3280*u**5*v**4 + 1768*u**5*v**3 - 1824*u**5*v**2 + 1336*u**5*v + 4680*u**5 + 72*u**4*v**6 + 760*u**4*v**5 + 2972*u**4*v**4 + 5312*u**4*v**3 + 5212*u**4*v**2 + 5800*u**4*v + 5480*u**4 + 128*u**3*v**5 + 1088*u**3*v**4 + 3456*u**3*v**3 + 5632*u**3*v**2 + 6336*u**3*v + 4608*u**3 + 144*u**2*v**4 + 960*u**2*v**3 + 2448*u**2*v**2 + 3360*u**2*v + 2448*u**2 + 96*u*v**3 + 480*u*v**2 + 896*u*v + 736*u + 32*v**2 + 96*v + 96
    print('h(u<=1,v<=1) =', factor(h.subs(u, 1/(1 + u)).subs(v, 1/(1 + v))))
    print('h(u<=1,v>=1) =', factor(h.subs(u, 1/(1 + u)).subs(v, 1 + v)))
    print('h(u>=1,v<=1) =', factor(h.subs(u, 1 + u).subs(v, 1/(1 + v))))
    print('h(u>=1,v>=1) =', factor(h.subs(u, 1 + u).subs(v, 1 + v)))

if __name__ == '__main__':
    main()
```
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  • $\begingroup$ Yes, BW works. Sometimes, we can do BW by hand when it is not very complicated. Otherwise, doing by hand is prohibited. $\endgroup$
    – River Li
    Oct 9, 2023 at 2:11

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