Algebraic inequality $\sum \frac{x^3}{(x+y)(x+z)(x+t)}\geq \frac{1}{2}$ The inequality is
$$\frac{x^3}{(x+y)(x+z)(x+t)}+\frac{y^3}{(y+x)(y+z)(y+t)}+\frac{z^3}{(z+x)(z+y)(z+t)}+\frac{t^3}{(t+x)(t+y)(t+z)}\geq \frac{1}{2},$$
for $x,y,z,t>0$.
It originates from a 3-D geometry problem involving volumes of tetrahedra etc. Actually, it is equivalent with that problem (see Let ABCD be a tetrahedron of volume 1 and M,N,P,Q,R,S on AB,BC,CD,DA,AC,BD s.t. MP,NQ,RS are concurrent. Then the volume of MNRSPQ is less than 1/2.).
The three variables simpler case
$$\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+x)(y+z)}+\frac{z^2}{(z+x)(z+y)}\geq \frac{3}{4}$$
can be proved using the Cauchy-Schwarz inequality in "Engel's form".
I have tried variants of Holder type inequalities, until now unsuccessfully.

As inequality homogenous WLOG $x+y+z+t=4$
By AM-GM$$\sum \frac{x^3}{(x+y)(x+t)(x+z)} \ge\sum \frac{27x^3}{{(3x+y+z+t)}^3}= \sum \frac{27x^3}{{(2x+4)}^3}$$  By tangent line method $$\frac{x^{3}}{\left(2x+4\right)^{3}}\ge \frac{1}{6^{3}}+\frac{1}{108}\left(x-1\right)$$ $$\iff -\frac{\left(x-1\right)^{2}\left(x^{2}-6x-4\right)}{108\left(x+2\right)^{3}}\ge 0$$ which is true as $x^2-6x-4<0,\forall x\in[0,4]$ So $$  \sum \frac{27x^3}{{(2x+4)}^3}\ge \frac{27 \cdot 4}{6^3}+\frac{0}{108}=\frac{1}{2}$$ done!!
