Why can $\frac{{{x^2}}}{{\sqrt y }} \le t$ be prepresent as the the following equivalent second order cone constraint?

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I am quite new to the field of convex optimization and in a research paper that I have read, some author represent this constraint $\frac{{{x^2}}}{{\sqrt y }} \le t$ as equivalent SOC constraint like this without explanation

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Could you kindly explain to me why this is possible and how can they come up with an excellent use of the auxiliary variable $v$ to turn the original constrain into SOC form ?

Thank you for your enthusiasm !

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That's because you can use SOC to write a constraint $$pq\geq r^2,$$ namely as $$\|2r,p-q\|_2\leq p+q,$$ and that makes it possible to write various simple inequalities between monomials, and then put them together to get more complicated monomials.

Since you want to model $t\sqrt{y}\geq x^2$ it feels natural to start with $$tv\geq x^2$$ and to complete the model you are forced to take $$\sqrt{y}\geq v,$$ which is also fine because it is equivalent to the same type of SOC: $$1\cdot y\geq v^2.$$

These are rather well-known tricks you can read about in https://docs.mosek.com/modeling-cookbook/cqo.html#simple-sets-involving-power-functions.

PS. The second inequality from your post looks wrong, $v$ and $y$ should be switched.