Cauchy–Schwarz type Inequality for 4 variables 
For $x,y,z,t \geqslant 0$, prove that
  $$ \tag{1} (x+2y+3z+4t)(x^2+y^2+z^2+t^2) \geqslant \frac{35-\sqrt{10}}{54}(x+y+z+t)^3$$

Observations


*

*This inequality is not symmetric nor cyclic. We cannot order the variables.

*This has 4 variables. My first attempt is to assume that $x+y+z+t=1$, then plug in $t=1-x-y-z$ to $(1)$ and try to reduce number of variables. But there are so many cases to consider.

*I think Cauchy–Schwarz is a good one. I just don't know how to get the cosntant $\frac{35-\sqrt{10}}{54}$ .


This is a typical problem that can be solved by brute force with homogeneization techniques and Lagrange multipliers. Since the inequality is homogeneous, it is not restrictive to assume $x^2+y^2+z^2+t^2=1$, then compute
$$ \max_{x^2+y^2+z^2+t^2=1} f(x,y,z,t)=\max_{x^2+y^2+z^2+t^2=1}\frac{x+2y+3z+4t}{(x+y+z+t)^3} $$
by imposing $2x = \lambda\frac{\partial f}{\partial x}$ and so on. Lagrange equations then give that the stationary points are eigenvectors for the rank-$2$ matrix
$$ M=\begin{pmatrix}2&5&8&11\\ 1&4&7&10 \\ 0&3&6&9 \\ -1&2&5&8 \end{pmatrix}$$
that has a unique eigenvector with positive coordinates. It follows that the maximum is achieved at
$$ (x,y,z,t) = \mu\left(\sqrt{10}-1,\frac{2}{3}\sqrt{10}-\frac{1}{3},\frac{1}{3}\sqrt{10}+\frac{1}{3},1\right) $$
and $\mu$ can be computed by imposing that $x^2+y^2+z^2+t^2=1$. That also proves
$$\max_{x^2+y^2+z^2+t^2=1} f(x,y,z,t)=\frac{35-\sqrt{10}}{54}$$
as wanted.
