Let $x,y,z,w>0$ and $x+y+z+w=1$. What is the minimum value of $$\frac{x^2+y^2+z^2+w}{(x+y+z)^3}+\frac{y^2+z^2+w^2+x}{(y+z+w)^3}+\frac{z^2+w^2+x^2+y}{(z+w+x)^3}+\frac{w^2+x^2+y^2+z}{(w+x+y)^3}?$$
When all variables are equal (to $0.25$), the value of the expression is $4\frac{4}{27}$. If we want to homogenize the numerator, we can do $$x^2+y^2+z^2+w=x^2+y^2+z^2+w^2+wx+wy+wz.$$ Applying the AM-GM inequality then gives $$x^2+y^2+z^2+w\geq \frac{2}{3}(xy+yz+zx)+\frac{5}{3}(wx+wy+wz).$$
Yes, you are right! The minimal value is $4\frac{4}{27}.$
Indeed, by the Cauchy-Schwarz Inequality we obtain: $$\sum_{cyc}\frac{w}{x+y+z}=\sum_{cyc}\frac{x+y+z+w}{x+y+z}-4=$$ $$=\frac{1}{3}\sum_{cyc}(x+y+z)\sum_{cyc}\frac{1}{x+y+z}-4\geq\frac{16}{3}-4=\frac{4}{3}.$$ That is, by Cauchy-Schwarz again we obtain: $$\sum_{cyc}\frac{x^2+y^2+z^2+w}{(x+y+z)^3}=\sum_{cyc}\frac{x^2+y^2+z^2}{(x+y+z)^3}+\sum_{cyc}\frac{w^2+w(x+y+z)}{(x+y+z)^3}\geq$$ $$\geq\frac{1}{3}\sum_{cyc}\frac{(x+y+z)^2}{(x+y+z)^3}+\sum_{cyc}\frac{w^2(x+y+z+w)}{(x+y+z)^3}+\sum_{cyc}\frac{w(x+y+z+w)}{(x+y+z)^2}=$$ $$=\frac{1}{3}\sum_{cyc}\frac{x+y+z+w}{x+y+z}+\sum_{cyc}\frac{w^3}{(x+y+z)^3}+2\sum_{cyc}\frac{w^2}{(x+y+z)^2}+\sum_{cyc}\frac{w}{x+y+z}=$$ $$=\frac{4}{3}+\frac{4}{3}\sum_{cyc}\frac{w}{x+y+z}+\sum_{cyc}\frac{w^3}{(x+y+z)^3}+2\sum_{cyc}\frac{w^2}{(x+y+z)^2}\geq$$ $$\geq\frac{4}{3}+\frac{4}{3}\sum_{cyc}\frac{w}{x+y+z}+\frac{1}{16}\left(\sum_{cyc}\frac{w}{x+y+z}\right)^3+\frac{1}{2}\left(\sum_{cyc}\frac{w}{x+y+z}\right)^2\geq$$ $$\geq\frac{4}{3}+\frac{4}{3}\cdot\frac{4}{3}+\frac{1}{16}\left(\frac{4}{3}\right)^3+\frac{1}{2}\left(\frac{4}{3}\right)^2=\frac{112}{27}.$$