Given the portion of the sphere which is in the first octant: $x^2+y^2+z^2=1$ ; $ x,y,z>0$ , show the following cyclic inequality:
$$ \sum_{cyc} \frac{x^2+x}{x^2 +yz} = \frac{x^2+x}{x^2 +yz}+ \frac{y^2+y}{y^2 +xz} +\frac{z^2+z}{z^2 +xy} \ge 4 $$
Equality occurs in the limit $(1,0,0)$ and permutations thereof.
I have used 3d-spherical coordinates and the resulting function of $\theta$ and $\phi$ exhibits the inequality graphically.
I have also tried the following, using Cauchy-Schwarz and the spherical condition, which does not work (to save pain...):
$$ \sum_{cyc} \frac{x^2}{x^2 +yz} \ge \frac{(\sum_{cyc} x)^2}{\sum_{cyc} (x^2 +yz) } = \frac{1 + 2 \sum_{cyc} yz}{1 + \sum_{cyc} yz } = \frac{ \sum_{cyc} yz}{1 + \sum_{cyc} yz } + 1 $$ Being very brave, it would then be enough to prove $$ \sum_{cyc} \frac{x}{x^2 +yz} \ge 3 $$ however numerics show that this does not hold. Being not so brave it would be enough to prove $$ \sum_{cyc} \frac{x}{x^2 +yz} + \frac{ \sum_{cyc} yz}{1 + \sum_{cyc} yz } \ge 3 $$ and I have no clue on that one.