# $3\geq\sum\limits_{cyc}\frac{(x+y)^{2}x^{2}}{(x^{2}+y^{2})^{2}}$ with $x,y,z >0$

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.

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

## 2 Answers

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 $$F(x)$$:

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.

• 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 at 12:10
• @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 at 12:21
• Yes, good work, 1+ – dan_fulea Feb 11 at 10:41

## 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 : $$0 and $$0 and $$0 and $$\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 : $$0 and $$0 and $$0 and $$\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 ?

• 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 at 12:58
• @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 . – Erik Satie Jan 31 at 15:18
• @RiverLi Please check my second edit .Thanks again. – Erik Satie Jan 31 at 16:14
• 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 1 at 1:31
• @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. – Erik Satie Feb 1 at 14:20