<s>[INST] Minimize $\frac{(x^2+1)(y^2+1)(z^2+1)}{ (x+y+z)^2}$, $x,y,z>0$ Minimize $\;\;\displaystyle \frac{(x^2+1)(y^2+1)(z^2+1)}{ (x+y+z)^2}$, if $x,y,z>0$. 
By setting gradient to zero I found $x=y=z=\frac{1}{\displaystyle\sqrt{2}}$, which could minimize the function.

Question from Jalil Hajimir

 [/INST] If you want some calculus/analysis argument:
After establishing there must exist a global minimum, let $p$ be the global minimum. Then we must have that
$$f(x) = x^2\left((y^2+1)(z^2+1) - p\right) - 2xp(y+z) + (y^2+1)(z^2+1) - p(y+z)^2\geq 0$$
as a quadratic in $x.$ So the discriminant is non-positive:
$$D =4\left[p^2(y+z)^2 - (y^2+1)^2(z^2+1)^2 - p^2(y+z)^2+(y^2+1)(z^2+1)p(1+(y+z)^2)\right]\leq 0\iff $$
$$p\leq\min\dfrac{(y^2+1)(z^2+1)}{1+(y+z)^2}.$$
But
$$4(y^2+1)(z^2+1) - 3 - 3(y+z)^2 = 4y^2z^2+y^2+z^2-6yz+1 = (y-z)^2+(2yz-1)^2\geq 0.$$ So $p = \dfrac{3}{4}$ by continuity argument and it is achieved by $y = z = \dfrac{1}{\sqrt{2}},$ which in return easily tells us that that $x$ is also $\dfrac{1}{\sqrt{2}}$ for the minimum to be attained. 
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