Baltic Way shortlist

Let $x,y,z,t$ be positive numbers such that $a=x^4,b=y^4,c=z^4,d=t^4$. By AM-GM inequality $x^4+y^4+z^4+1\ge 4xyz$, $y^4+z^4+1+1\ge 4yz$ and $z^4+1+1+1\ge 4z$. Therefore we have the following estimation for the first fraction $$\frac{1}{\sqrt{x^4+2y^4+3z^4+10}}\le \frac{1}{\sqrt{4xyz+4yz+4z+4}}=\frac{1}{2\sqrt{xyz+yz+z+1}}.$$ Transform analogous estimations for the other fractions: $$\begin{array}{rl}\frac{1}{\sqrt{b+2c+3d+10}}& \le \frac{1}{2\sqrt{yzt+zt+t+1}}=\frac{1}{2\sqrt{t\sqrt{yz+z+1+xyz}}}=\frac{\sqrt{xyz}}{2\sqrt{xyz+yz+z+1}};\\ \frac{1}{\sqrt{c+2d+3a+10}}& \le \frac{1}{2\sqrt{ztx+tx+x+1}}=\frac{1}{2\sqrt{tx\sqrt{z+1+xyz+yz}}}=\frac{\sqrt{yz}}{2\sqrt{xyz+yz+z+1}};\\ \frac{1}{\sqrt{d+2a+3b+10}}& \le \frac{1}{2\sqrt{txy+xy+y+1}}=\frac{1}{2\sqrt{txy\sqrt{1+xyz+yz+z}}}=\frac{\sqrt{z}}{2\sqrt{xyz+yz+z+1}}.\end{array}$$ Thus, the sum does not exceed $$\frac{1+\sqrt{xyz}+\sqrt{yz}+\sqrt{z}}{2\sqrt{xyz+yz+z+1}}.$$ It remains to apply inequality $\sqrt{\alpha }+\sqrt{\beta }+\sqrt{\gamma }+\sqrt{\delta }\le 2\sqrt{\alpha +\beta +\gamma +\delta }$, which can be easily proven by taking squares or derived from inequality between arithmetical and quadratic means.