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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.

Michael Rozenberg
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NKellira
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3 Answers3

5

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.

River Li
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    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 $\ne 0$. – dan_fulea Feb 10 '20 at 12:10
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    @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. – River Li Feb 10 '20 at 12:21
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    Yes, good work, 1+ – dan_fulea Feb 11 '20 at 10:41
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    Do you have any stronger form for this inequality? – NKellira Feb 26 '21 at 14:21
  • @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 $C$. – River Li Feb 26 '21 at 15:10
  • I can't, the degree is high. I hope that there is an elementary solution for $3\geq \frac{(x+ y)^{2}x^{2}}{(x^{2}+ y^{2})^{2}}+ \frac{(y+ z)^{2}y^{2}}{(y^{2}+ z^{2})^{2}}+ \frac{(z+ x)^{2}z^{2}}{(z^{2}+ x^{2})^{2}}$ – NKellira Feb 26 '21 at 23:11
  • @tthnew I hope so. – River Li Feb 26 '21 at 23:43
1

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 ?

Barackouda
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    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? – River Li Jan 31 '20 at 12:58
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    @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 . – Barackouda Jan 31 '20 at 15:18
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    @RiverLi Please check my second edit .Thanks again. – Barackouda Jan 31 '20 at 16:14
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    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? – River Li Feb 01 '20 at 01:31
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    @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. – Barackouda Feb 01 '20 at 14:20
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    The restriction $abc=1, a,b,c \in (0.67, 1.22)$ is quite strict. By the way, one well-known method for this type of problems here is to prove that $1-(1+x)^2/(x^2+1)^2- \ln(x)\ge 0$ for $x\in (0, 1.5511)$. Thus, the inequality is true for $abc=1, a, b, c \in (0, 1.5511)$. – River Li Feb 01 '20 at 14:59
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    @RiverLi Using $\cos^{-1}(x)$ I have a similar result for $a,b,c \in(0,1.45)$ .The two advantages of Jensen's inequality is the refinement in this case and we can avoid derivative using the definition of convexity .I was not aware of this method have you a link ?Thanks . – Barackouda Feb 01 '20 at 15:36
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    https://math.stackexchange.com/questions/554121/how-prove-this-inequality-a2b2c28abbcac3-10abc-ge-0,
    https://math.stackexchange.com/questions/1000997/how-to-prove-the-inequality-fraca-sqrt1a-fracb-sqrt1b-fracc, https://artofproblemsolving.com/community/c6h207944p1144256, https://artofproblemsolving.com/community/c6h195117p1071741, https://artofproblemsolving.com/community/c6h1510875p8968214, https://artofproblemsolving.com/community/c6h1588979p9841359     – River Li Feb 01 '20 at 16:15
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    @RiverLi ok I understand it's very simple since $\ln(abc)=0$ .Can we show the inequality if we study the inequality $1-(1+x)^2/(1+x^2)^2-\alpha\ln(x)\geq 0$ where $\alpha>0$ is a parameter ? – Barackouda Feb 02 '20 at 13:53
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    Yes, denote it by $g(x)$, and you just let $g'(1) = 0$ to determine $\alpha$. – River Li Feb 02 '20 at 14:06
1

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*x2/(x2 + y2)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*u12 + 20u11v + 40u11 + 43u10*v2 + 172u10v + 192u10 + 54u9*v3 + 302u9v2 + 692*u9v + 576u9 + 47*u8v4 + 286u8*v3 + 935u8v2 + 1700*u8v + 1188u8 + 32*u7v5 + 174u7*v4 + 548u7v3 + 1624*u7v2 + 2816u7v + 1752u7 + 17u6v6 + 88*u6v5 + 96u6*v4 + 228u6v3 + 1688*u6v2 + 3296u6v + 1864u6 + 6u5v7 + 42*u5v6 - 10u5*v5 - 524u5v4 - 728*u5v3 + 1088u5*v2 + 2792u5v + 1408u5 + u4v8 + 14*u4v7 + 20u4*v6 - 276u4v5 - 1080*u4v4 - 1152u4*v3 + 588u4v2 + 1728*u4v + 720u4 + 2*u3v8 + 16u3*v7 - 4u3v6 - 248*u3v5 - 640u3*v4 - 416u3v3 + 512*u3v2 + 768u3v + 224u3 + 3u2v8 + 24*u2v7 + 84u2*v6 + 160u2v5 + 252*u2v4 + 384u2*v3 + 432u2v2 + 224*u2v + 32u2 + 4uv8 + 40uv7 + 152uv6 + 328uv5 + 448uv4 + 384uv3 + 192uv2 + 32uv + 4v8 + 24v7 + 72*v6 + 128v5 + 144v4 + 96*v3 + 32v2
    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 = u10v8 + 14*u10v7 + 87u10*v6 + 316u10v5 + 742*u10v4 + 1168u10*v3 + 1216u10v2 + 768*u10v + 224u10 + 2*u9v8 + 30u9*v7 + 196u9v6 + 746*u9v5 + 1874u9*v4 + 3304u9v3 + 4064*u9v2 + 3136u9v + 1120u9 + 3u8v8 + 40*u8v7 + 216u8*v6 + 614u8v5 + 1116*u8v4 + 1960u8*v3 + 3775u8v2 + 4796*u8v + 2492u8 + 4*u7v8 + 56u7*v7 + 276u7v6 + 428*u7v5 - 844u7*v4 - 3644u7v3 - 3040*u7v2 + 2332u7v + 3352u7 + 4u6v8 + 72*u6v7 + 476u6*v6 + 1320u6v5 + 620*u6v4 - 4224u6*v3 - 7244u6v2 - 736*u6v + 3764u6 + 24*u5v7 + 320u5*v6 + 1576u5v5 + 3280*u5v4 + 1768u5*v3 - 1824u5v2 + 1336*u5v + 4680u5 + 72*u4v6 + 760u4*v5 + 2972u4v4 + 5312*u4v3 + 5212u4*v2 + 5800u4v + 5480u4 + 128u3*v5 + 1088u3v4 + 3456*u3v3 + 5632u3*v2 + 6336u3v + 4608u3 + 144u2*v4 + 960u2v3 + 2448*u2v2 + 3360u2v + 2448u2 + 96uv3 + 480uv2 + 896uv + 736u + 32v*2 + 96v + 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()
```
auntyellow
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  • Yes, BW works. Sometimes, we can do BW by hand when it is not very complicated. Otherwise, doing by hand is prohibited. – River Li Oct 09 '23 at 02:11