Determine the minimum of:$\frac{\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}}{a+b+c+d},\text{ if }a,b,c,d>0$ Determine the minimum of:$\frac{\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}}{a+b+c+d}$, if $a,b,c,d>0$ I tried partial derivatives of the unction and also its natural log aI came up with system of equation that is not easy to solve. Using AM GM easily you can show the given fraction is greater or equal than 1

Let $a=\frac{x}{\sqrt3},$ $b=\frac{y}{\sqrt3},$ $c=\frac{z}{\sqrt3}$ and $d=\frac{t}{\sqrt3}$.

Thus, by C-S we obtain: $$\frac{\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}}{a+b+c+d}=\frac{\sqrt{(x^2+3)(y^2+3)(z^2+3)(t^2+3)}}{9(a+b+c+d)}\geq$$ $$\geq\frac{\sqrt{2((x+y)^2+4)\cdot2(4+(z+t)^2)}}{9(a+b+c+d)}\geq\frac{2(2(x+y)+2(y+t))}{9(a+b+c+d)}=\frac{4}{3\sqrt3}.$$ The equality occurs for $a=b=c=d=\frac{1}{\sqrt3},$ which says that we got a minimal value.

I used the following inequality: $$(x^2+3)(y^2+3)\geq2((x+y)^2+4),$$ which is $$(xy-1)^2+(x-y)^2\geq0.$$