<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] We need to prove that:
$$2\sum_{cyc}x^3(y+z)(y+t)(z+t)\geq(x+y)(x+z)(x+t)(y+z)(y+t)(z+t)$$ or
$$2\sum_{sym}\left(x^3y^2z+\frac{1}{3}x^3yzt\right)\geq\sum_{sym}\left(x^3y^2z+\frac{1}{3}x^3yzt+x^2y^2zt+\frac{1}{3}x^2y^2z^2\right)$$ or
$$\sum_{sym}(3x^3y^2z-x^2y^2z^2+x^3yzt-3x^2y^2zt)\geq0,$$ which is true by Muirhead because
$(3,2,1,0)\succ(2,2,2,0),$ $(3,2,1,0)\succ(2,2,1,1)$ and $(3,1,1,1)\succ(2,2,1,1).$
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