Does the following series of transformations of inequalities holds?

Question

I am to calculate limit of the function $f(x,y)$ i am trying to apply squeeze theorem. Is the following series of transformations of this inequality correct? If not how to do this correctly? i.e. are those inequalities true or false? They should hold for all positive numbers. $$f(x,y)=\frac{\ln \left(1+x^3+y^3 \right)}{\sqrt{x^2+y^2}} = \frac{\sqrt{x^2+y^2} \ln \left(1+x^3+y^3 \right)}{x^2+y^2} \le \frac{\sqrt{x^2+y^2} \left(x+y \right)}{x^2+y^2} \le \frac{\sqrt{x^2+y^2} }{1}$$

Answer 1

Note that your last inequality is equivalent to the inequality \begin{equation} x + y \leq x^2 + y^2 \quad \forall x, y > 0, \end{equation} which is false (e.g. take $x, y = \frac{1}{2}$).

Perhaps use the following:

• For any $\alpha > 0$, $\ln(1 + \alpha) \leq \alpha$.
• $x^3 + y^3 \leq (x^2 + y^2)(x + y) \leq (x^2 + y^2)^{3/2}$, provided $x, y > 0$.

You should find $f(x,y) \leq x^2 + y^2$, $\forall x, y > 0$.