Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3$ Prove that if $a,b,c,d$ are positive reals we have:
$$\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3.$$
I think that I have found a equality case, for example, when $a=x^4,b=x^3,c=x^2,d=x$, and as $x$ tends to $\infty$, the LHS tends to $3$, but this means that the inequality is very unlikely to be solved with traditional methods, such as Cauchy-Schwartz (my starting idea), so I got stuck.
 A: We first prove the following lemma.
Lemma. Let $\lambda>0$. Then
$$\sup_{x>0}\left(\frac1{\sqrt{1+x}}+\frac1{\sqrt{1+\frac{\lambda}x}}\right)=\begin{cases}\frac2{\sqrt{1+\sqrt\lambda}}&\text{if }\lambda<4\\\sqrt{\frac{\lambda}{\lambda-1}}&\text{if }\lambda\geq 4.\end{cases}$$
Proof. Fixing $\lambda$ constant, let the left side be $f(x)$. The function $f$ is clearly continuous and infinitely differentiable, and so its global extrema are reached when $x\to 0$, $x\to\infty$, or when $f''(x)=0$. It is not hard to compute that
$$f''(x)=0\Longleftrightarrow \lambda^2(1+x)^3=x(\lambda+x)^3;$$
the polynomial on the right factors as $(x^2-\lambda)(x^2-(\lambda^2-3\lambda)x+\lambda)$. So, the extrema are when $x=\sqrt{\lambda}$, or when $x$ and $\lambda/x$ satisfy $x+\frac\lambda x=\lambda^2-3\lambda$. For $\lambda<4$, $x+\frac\lambda x$ is always larger than $\lambda^2-3\lambda$, so the only $x$ with $f''(x)=0$ is $x=\sqrt\lambda$, giving the desired bound. For $\lambda\geq 4$, such a real $x>0$ does exist, and satisfies, letting $y=\lambda/x$ so that $x+y=\lambda^2-3\lambda$,
\begin{align*}
\left(\frac1{\sqrt{1+x}}\right)\left(\frac1{\sqrt{1+y}}\right)&=\frac{1}{\sqrt{1+(\lambda^2-3\lambda)+\lambda}}=\frac1{\lambda-1}\\
\left(\frac1{\sqrt{1+x}}\right)^2+\left(\frac1{\sqrt{1+y}}\right)^2&=\frac{2+x+y}{(1+x)(1+y)}=\frac{\lambda-2}{\lambda-1},
\end{align*}
so
$$\frac1{\sqrt{1+x}}+\frac1{\sqrt{1+y}}=\sqrt{\frac{\lambda-2}{\lambda-1}+\frac2{\lambda-1}}=\frac{\lambda}{\lambda-1}.$$
For $\lambda\geq 4$, this is always at least the value $\frac2{\sqrt{1+\sqrt\lambda}}$ achieved at $x=\sqrt\lambda$, and so is the maximum of $f$. $\square$

Now, under the substitution $w=\frac{b^2}{a^2}$, $x=\frac{c^2}{b^2}$, $y=\frac{d^2}{c^2}$, $z=\frac{a^2}{d^2}$, the problem becomes to show
$$\frac1{\sqrt{1+w}}+\frac1{\sqrt{1+x}}+\frac1{\sqrt{1+y}}+\frac1{\sqrt{1+z}}\leq 3$$
for $w,x,y,z>0$ and $wxyz=1$. If $wx,yz\leq 4$, then the sum is at most
$$\frac2{\sqrt{1+t}}+\frac2{\sqrt{1+\frac1t}}\leq \frac4{\sqrt{2}}=2\sqrt2<3$$
for $t=\sqrt{wx}$, where we have used our lemma on $\lambda=t^2=wx$, $\lambda=1/t^2=yz$, and again on $\lambda=1$. Otherwise, assume without loss of generality that $wx\geq 4$, and let $s=\sqrt{wx}$. Then, by our lemma, the sum is at most
$$\sqrt{\frac{s^2}{s^2-1}}+\frac{2}{\sqrt{1+\frac1s}},$$
where $s\geq 2$. However, for $s\geq 2$,
\begin{align*}
\sqrt{\frac{s^2}{s^2-1}}+\frac{2}{\sqrt{1+\frac1s}}
&=\frac{\sqrt s(\sqrt s+2\sqrt{s-1})}{\sqrt{s^2-1}}\\
&\leq \frac{\sqrt s\left(\sqrt s+2\left[\sqrt s-\frac1{2\sqrt s}\right]\right)}{\sqrt{s^2-1}}\\
&=\frac{3s-1}{\sqrt{s^2-1}},
\end{align*}
which is at most $3$ whenever $s\geq 5/3$ since
$$(3s-1)^2=9s^2-6s+1\leq 9s^2-9\implies \frac{3s-1}{\sqrt{s^2-1}}\leq 3.$$
A: Since the inequality is both homogeneous and cyclic, assume that $a = \max(a, b, c, d) = 1$.
It suffices to prove that, for all $b, c, d \in (0, 1]$,
$$\frac{1}{\sqrt{1+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+1}} \le 3.$$
To this end, first, we have
\begin{align*}
 \frac{1}{\sqrt{1+b^2}}+\frac{b}{\sqrt{b^2+c^2}}
 &= \sqrt{\frac{1}{1+b^2} + \frac{b^2}{b^2+c^2} + \frac{2b}{\sqrt{(1 + b^2)(b^2 + c^2)}}}\\
 &\le \sqrt{\frac{1}{1+b^2} + \frac{b^2}{b^2+c^2} + \frac{2b}{b + bc}}\\
 &= \sqrt{\frac{1}{1+b^2} + \frac{b^2}{b^2+c^2} + \frac{2}{1 + c}}\\
 &\le 1 + \frac14 \left(\frac{1}{1+b^2} + \frac{b^2}{b^2+c^2} + \frac{2}{1 + c}\right)\\
 &\le 1 + \frac14\left(\frac{2}{1 + c} + \frac{2}{1 + c}\right)\\
 &= 1 + \frac{1}{1 + c} \tag{1}
\end{align*}
where we have used Cauchy-Bunyakovsky-Schwarz inequality,
and $\sqrt{u} \le 1 + \frac{u}{4}$,
and $\frac{2}{1+c} - \frac{1}{1+b^2} - \frac{b^2}{b^2+c^2} = \frac{(1-c)(b^2-c)^2}{(1+c)(b^2+1)(b^2+c^2)}\ge 0$.
Second, we have
\begin{align*}
 \frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+1}} &= \sqrt{\frac{c^2}{c^2+d^2}
  + \frac{d^2}{d^2+1} + \frac{2cd}{\sqrt{(c^2+d^2)(d^2+1)}}}\\
 &\le \sqrt{\frac{c^2}{c^2+d^2}
  + \frac{d^2}{d^2+1} + \frac{2cd}{cd + d}}\\
 &= \sqrt{\frac{c^2}{c^2+d^2}
  + \frac{d^2}{d^2+1} + \frac{2c}{c + 1}}\\
 &\le \frac12 + \frac12 \left( \frac{c^2}{c^2+d^2}
 + \frac{d^2}{d^2+1} + \frac{2c}{c + 1}\right)\\
 &\le \frac12 + \frac12 \left( \frac{1}{1+d^2}
 + \frac{d^2}{d^2+1} + \frac{2c}{c + 1}\right)\\
 &= \frac12 + \frac12 \left( 1 + \frac{2c}{c + 1}\right)\\
 &= 1 + \frac{c}{c + 1} \tag{2}
\end{align*}
where we have used Cauchy-Bunyakovsky-Schwarz inequality,
and $\sqrt{u} \le \frac12 + \frac12 u$.
Using (1) and (2), the desired result follows.
We are done.
A: Partial hint :
We have a two first  not hard to show :
Let $x>0$ then define :
$$f\left(x\right)=\frac{x}{\sqrt{x^{2}+1}}$$
Then :
$$f''(x)\leq 0$$
Second fact :
Let $x>0$ then define :
$$(f(e^x))''<0$$
So we use Jensen's inequality with the constraint $c\geq d\ge a\geq 1$ we need  to show for $x,y\leq 1$ :
$$2f\left(\frac{\left(x+y\right)}{2}\right)+2f\left(\sqrt{z}\right)\le3$$
Where $x=\frac{a}{b},y=\frac{b}{c},z=\frac{c}{a}$ and $xyz=1$
Wich is easier and true.


We have the easy inequalities firstly for $0<x\leq y$ :
$$\frac{\sqrt{2}xy}{x^{2}+y^{2}}-\frac{x}{\sqrt{x^{2}+y^{2}}}\geq 0$$
And for $0<y\leq x$ :
$$\frac{\sqrt{2}x^{2}}{x^{2}+y^{2}}-\frac{x}{\sqrt{x^{2}+y^{2}}}\geq 0$$
So assuming $y\geq z\geq  x> 0$ and $d\geq 0$ we have the inequality :
$$g\left(x,y\right)+g\left(y,z\right)+h\left(z,d\right)+h\left(d,x\right)\leq 3$$
$$g\left(x,y\right)=\frac{\sqrt{2}xy}{x^{2}+y^{2}}\quad\quad,h\left(x,y\right)=\frac{x}{\sqrt{x^{2}+y^{2}}}$$
A: Partial Hint :
For $x,y>0$ and $x,y\leq 1$ such that $\left(y\right)^{2}+\left(x-\frac{1}{5}\right)^{2}\geq\frac{2}{50}$ we have the inequality :
$$\sqrt{\frac{x^{2}}{y^{2}+x^{2}}}-\left(f\left(x\right)-f\left(y\right)\right)-\frac{3}{4}\le0$$
Where :
$$f\left(x\right)=x\sqrt{\frac{1}{x^{2}+1}}$$
Summing gives the inequality
Edit : we supposed that for :
$$h\left(x,y\right)=\frac{3x^{2}}{2\left(x^{2}+y^{2}\right)} $$
We have :
$$h\left(a,b\right)+h\left(b,c\right)+h\left(c,d\right)+h\left(d,a\right)\leq3$$
Wich can be done since the inequality is homogeneous .
Now we have the inequality for $x,y>6$ and $x/5\leq y$  :
$$\sqrt{\frac{x^{2}}{x^{2}+y^{2}}}-f\left(x\right)+f\left(y\right)-\frac{3x^{2}}{2\left(x^{2}+y^{2}\right)}\le0$$
Where :
$$f\left(x\right)=0.5\ln\left(\sqrt{\frac{x}{x^{2}+1}}\right)$$
I haven't a proof yet but I come back again on it .
Hope it helps .
A: Again partial hint :
Using the formula :
$$\cos\left(\operatorname{arccot}\left(\frac{a}{b}\right)\right)-\sqrt{\frac{a^{2}}{a^{2}+b^{2}}}=0$$
We use Karamata's inequality with $-\cos(x)$:
$$\frac{a}{b}\leq \frac{b}{c}\leq \frac{c}{d}\leq \frac{d}{a}$$
And $$3=4\cos\left(\operatorname{arccot}\left(\frac{3}{\sqrt{7}}\right)\right)$$
With constraint :
$$f(a,b)\geq \operatorname{arccot}\left(\frac{3}{\sqrt{7}}\right),\quad,f(a,b)+f(b,c)\geq 2 \operatorname{arccot}\left(\frac{3}{\sqrt{7}}\right),f(a,b)+f(b,c)+f(c,d)\geq 3\operatorname{arccot}\left(\frac{3}{\sqrt{7}}\right),\quad f\left(a,b\right)+f\left(b,c\right)+f\left(c,d\right)+f\left(d,a\right)\geq4\operatorname{arccot}\left(\frac{3}{\sqrt{7}}\right)$$
Where :
$$f\left(x,y\right)=\operatorname{arccot}\left(\frac{x}{y}\right)$$
