<s>[INST] 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.
 [/INST] This is a solution using Carlson inequality: if $x_{ij}(1\leq i\leq m,1\leq j\leq n)$ are non-negative reals, then
$$
\prod_{s=1}^m\sum_{t=1}^nx_{st}\geq\left(\sum_{t=1}^n\left(\prod_{s=1}^mx_{st}\right)^{\frac{1}{m}}\right)^{m}
$$

Proof. According to the Carlson inequality,
$$
\begin{align}
&\sum\frac{x^3}{(x+y)(x+z)(x+t)}\cdot\sum(x+y)\cdot\sum(x+z)(x+t)\\
\geq&\left(\sum x\right)^3\\
\because&\sum(x+y)=2(x+y+z+t),\sum(x+z)(x+t)=(x+y+z+t)^2\\
\therefore&\sum\frac{x^3}{(x+y)(x+z)(x+t)}\cdot2(x+y+z+t)\cdot(x+y+z+t)^2\geq(x+y+z+t)^3
\end{align}
$$
One can now easily observe that the original inequality is true.
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