An inequality regarding calculus

Question

Let $f:\mathbb{R_+} \times \mathbb{R_+}\to \mathbb{R}$ with $$f(x,y)=\frac{(x^2+1)(y^2+1)}{x+y}$$ Prove that $$f(x,y) \geq \frac{8}{\sqrt{27}}, \: \forall x,y>0$$

The problem first suggests proving that $f(x,y)\geq f(\sqrt{y^2+1}-y,y)$, which I managed to do by straightforward substitution and also by derivatives. This led me to the inequality from here, but I'm wondering if there is another approach, which maybe does not involve much calculus.

Answer 1

solve the System $$y^2+2xy-1=0$$ $$x^2+2xy-1=0$$ this Comes from $$\frac{\partial f(x,y)}{\partial x}=0$$ $$\frac{\partial f(x,y)}{\partial y}=0$$

Answer 2

The function blows up to infinity if $x$ or $y$ get large, or if $x$ or $y$ approach $0$. Hence there is a minimizer in the interior of the orthant. Solving Dr. Sonnhard Graubner's equations give $x=y=1/\sqrt{3}$. Plugging this in the objective function gives the lower bound.