8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Proving any non-trivial inequality with the Buffalo way is not fun. $\endgroup$ – Display name Jan 29 at 1:23
  • $\begingroup$ 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. $\endgroup$ – Display name Jan 29 at 1:24
  • $\begingroup$ 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 $uvw=1.$ $\endgroup$ – Display name Jan 29 at 1:26
  • $\begingroup$ 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. $\endgroup$ – tthnew Jan 29 at 1:33
  • $\begingroup$ @Display name BW does not help here! The Vasc's theorems don't work! $\endgroup$ – Michael Rozenberg Jan 29 at 5:20
5
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 at 12:10
  • $\begingroup$ @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 at 12:21
  • $\begingroup$ Yes, good work, 1+ $\endgroup$ – dan_fulea Feb 11 at 10:41
2
$\begingroup$

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 ?

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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 at 12:58
  • 1
    $\begingroup$ @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 . $\endgroup$ – Erik Satie Jan 31 at 15:18
  • 1
    $\begingroup$ @RiverLi Please check my second edit .Thanks again. $\endgroup$ – Erik Satie Jan 31 at 16:14
  • 1
    $\begingroup$ 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 at 1:31
  • 1
    $\begingroup$ @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. $\endgroup$ – Erik Satie Feb 1 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.