<s>[INST] $\frac{1}{x+1}+\frac{1}{y+1} +\frac{1}{z+1}$ minimum value if $xyz$ =k.$x,y,z$ are positive reals. $\frac{1}{x+1}+\frac{1}{y+1} +\frac{1}{z+1}$ minimum value if $xyz$ =k.$x,y,z$ are positive reals.I think the minimum should be when $x=y=z=k^{1/3}$. How do I show it? I tried to use AM-GM inequality but it doesnt seem to work.
 [/INST] First, note that it is not true that always the function takes minimum when $x=y=z$.  For e.g., let $r = \sqrt[3]k, \; x = y = rt,\, z= \dfrac{r}{t^2}$ for $r, t>0$.  Then our objective function is
$$f(t) = \frac1{rt+1}+\frac1{rt+1}+\frac{t^2}{r+t^2}$$
Now consider $t \to \infty$.  Clearly $f(t) \to 1$.  Hence for $\dfrac3{r+1} > 1 \implies r < 2$, the minimum cannot be when $x=y=z$.
Consider now the case $r \ge 2$.  Further, let $x = r \dfrac{u}v, \, y = r \dfrac{v}w , \, z = r \dfrac{w}u $.  Now we will show that
$$\sum_{cyc} \frac1{x+1} = \sum_{cyc} \frac{v}{ru+v} \ge \frac3{r+1}$$
so indeed in these cases we have the minimum when $x=y=z$.
By Cauchy Schwarz inequality,
$$\sum_{cyc} \frac{v^2}{ruv+v^2} \ge \frac{(u+v+w)^2}{r(uv+vw+wu)+(u^2+v^2+w^2)}$$
So it is enough to show that
$$(r+1)(u+v+w)^2 \ge 3\left(r(uv+vw+wu)+(u^2+v^2+w^2) \right)$$ 
$$\iff (r-2)(u^2+v^2+w^2-uv-vw-wu) \ge 0$$
which obviously holds for $r\ge 2$.
</s>