Prove that $\displaystyle \sum_{cyc} \dfrac {1}{\sqrt{a+2b+3c+10}} \leq 1$ for $a, b, c, d \in \mathbb{R}^+$ which satisfies $abcd=1$.$\Big($$\displaystyle \sum_{cyc}$ is for 4-variables.$\Big)$
\begin{align}
&\text{let } a = \frac y x, b = \frac z y, c = \frac w z, d= \frac x w. \\
& \Rightarrow \text{ETS) } \displaystyle \sum_{cyc} \frac {\sqrt{xyzw}}{\sqrt{y^2zw+2z^2wx+3w^2xy+10xyzw}} \leq 1.
\end{align}
I was thinking about the best expression to use the Cauchy-Schwartz inequality.
Best Answer
Let us generalize once for all times this inequality.
We fix some $n\ge 2$.
Let $a=(1,2,\dots,(n-1),N)$ with $N=\frac 12n(n-1)$, so the sum of the components is $n^2$.
Let $x=(x_1,x_2,\dots,x_n)$ be a tuple of positive numbers so that their product is one.
The group of permutations acts on the components of $x$, and we let below only the subgroup $C$ generated by the cycle $(123\dots n)$ act, sums will be built below over it, we write a "$x$-cyclic" attribute under the sums. So the $a$-coefficients are kept in place, and we build the scalar products of the $a$ vector with each of the $n$ cyclically permuted $x$ vectors.
Then we have the inequalities: $$ \begin{aligned} &\left( \sum_{x\text{-cyclic}} \frac 1{\sqrt{x_1+2x_2+\dots + (n-1)x_{n-1}+N}} \right)^2 \\ &\qquad\le n \sum_{x\text{-cyclic}} \frac 1{x_1+2x_2+\dots + (n-1)x_{n-1}+N} \\ &\qquad\le 1\ . \end{aligned} $$ The first inequality is Cauchy-Schwarz.
Let us show the second one. It is convenient to replace the variables $x_1,x_2,\dots,x_n$ by $y_1^n,y_2^n,\dots,y_n^n$ with positive $y_1,y_2,\dots, y_n$ of product also one.
We have: $$ \begin{aligned} &x_1+2x_2+3x_3+\dots + (n-1)x_{n-1}+N \\[3mm] &\qquad=\;y_1^n+2y_2^n+3y_3^n+\dots + (n-1)y_{n-1}^n+N \\[3mm] &\qquad=\; (y_1^n+y_2^n+y_3+\dots + y_{n-1}^n+1)\\ &\qquad\qquad +( 1+y_2^n+y_3^n+\dots + y_{n-1}^n+1)\\ &\qquad\qquad\qquad +( 1+1+y_3^n+\dots + y_{n-1}^n+1)\\ &\qquad\qquad\qquad\qquad + \dots \\ &\qquad\qquad\qquad\qquad\qquad +( 1+1+1+\dots + 1+1) \\[3mm] & \qquad \ge \; n(y_1^n\cdot y_2^n \cdot y_3\cdot \dots \cdot y_{n-1}^n \cdot 1)^{1/n}\\ &\qquad\qquad +n( 1\cdot y_2^n \cdot y_3^n \cdot \dots \cdot y_{n-1}^n\cdot 1)^{1/n}\\ &\qquad\qquad\qquad +n( 1 \cdot 1 \cdot y_3^n\cdot \dots \cdot y_{n-1}^n \cdot 1)^{1/n}\\ &\qquad\qquad\qquad\qquad + \dots \\ &\qquad\qquad\qquad\qquad\qquad +n( 1\cdot 1\cdot 1 \cdot \dots \cdot 1\cdot 1)^{1/n} \\[3mm] & \qquad =n\cdot y_1y_2y_3\dots y_{n-1}\ +\ n\cdot y_2y_3\dots y_{n-1}\ +\ n\cdot y_3\dots y_{n-1}\ +\ \dots\ +\ n \\[3mm] & \qquad =: n\; P(y_1,y_2,\dots,y_n)\ . \end{aligned} $$ The last line is a definition of $P$. It does not depend on $y_n$, but soon we act cyclically, and observe that multipication by $y_n$ brings... $$ (y_1y_2y_3\dots y_{n-1}\ +\ y_2y_3\dots y_{n-1}\ +\ \dots\ +\ 1) \overset{\cdot y_n}\longrightarrow (1\ +\ y_2y_3\dots y_{n-1}y_n\ +\ \dots\ +\ y_n) $$ so $$ P(y_1,y_2,\dots,y_n) \overset{\cdot y_n}\longrightarrow P(y_2,\dots,y_n,y_1) \overset{\cdot y_1}\longrightarrow P(y_3,\dots,y_1,y_2) \overset{\cdot y_2}\longrightarrow \dots\qquad . $$ So $$ \begin{aligned} &n\sum_{x\text{-cyclic}} \frac 1{x_1+2x_2+\dots + (n-1)x_{n-1}+N} \\ &\qquad = n\sum_{y\text{-cyclic}} \frac 1{y_1^n+2y_2^n+\dots + (n-1)y_{n-1}^n+N} \\ &\qquad\le n\sum_{y\text{-cyclic}} \frac 1{n\cdot P(y_1,y_2,\dots,y_n)} \\ &\qquad = \frac 1{P(y_1,y_2,\dots,y_n)} + \frac 1{P(y_2,y_3,\dots,y_1)} + \dots + \frac 1{P(y_n,y_1,\dots,y_{n-1})} \\ &\qquad = \frac 1{P(y_1,y_2,\dots,y_n)} + \frac 1{y_n\;P(y_1,y_2,\dots,y_n)} + \dots + \frac 1{y_ny_1\dots y_{n-2}\; P(y_1,y_2,\dots,y_n)} \\ &\qquad = \frac 1{P(y_1,y_2,\dots,y_n)}\left[ 1+\frac 1{y_n}+\dots+\frac 1{y_ny_1\dots y_{n-2}} \right] \\ &\qquad = \frac 1{P(y_1,y_2,\dots,y_n)} \left[ 1+y_1y_2\dots y_{n-1}+\dots+y_{n-1} \right] \\ &\qquad = \frac {P(y_1,y_2,\dots,y_n)}{P(y_1,y_2,\dots,y_n)} \\ &\qquad = 1\ . \end{aligned} $$ $\square$
As in the AoPS link added in the comments, Hard cyclic inequality with square roots, we have the following instances of this generalization: $$ \begin{aligned} 1 &\ge \frac 1{\sqrt{x+3}} + \frac 1{\sqrt{y+3}}\ , \qquad xy=1\ ,\ x,y> 0\ ,\\ 1 &\ge \frac 1{\sqrt{x+2y+6}} + \frac 1{\sqrt{y+2z+6}}+ \frac 1{\sqrt{z+2x+6}}\ ,\qquad xyz=1\ ,\ x,y,z> 0\ ,\\ 1 &\ge \frac 1{\sqrt{x+2y+3z+10}} +\frac 1{\sqrt{y+2z+3t+10}} +\frac 1{\sqrt{z+2t+3x+10}} +\frac 1{\sqrt{t+2x+3y+10}}\ , \end{aligned} $$ where in the last case $ xyzt=1$, $x,y,z,t> 0$.