# Proving $|E| \leq \frac{1}{2}(\sqrt{t-1}\;n^{3/2} + n$) in a graph with no copies of $K_{2,t}$

Show that if an $$n$$-vertex graph $$G=(V,E)$$ has no copy of $$K_{2,t}$$ then: $$|E| \leq \frac{1}{2}(\sqrt{t-1}\;n^{3/2} + n)$$

I know how to prove it for $$n=2: \;$$ Dentoe $$|E|=m,$$ $$d(v)$$ the degree of $$v$$ for every $$v \in V$$. Then: $$n \cdot\binom{\frac{2m}{n}}{2} \leq \sum_{v \in V} \binom{d(v)}{2}= \#K_{1,2} \leq \binom{n}{2}$$

The first inequality follows from the fact that the function $$x \to \binom{x}{2}$$ is convex and using jensen's inequality, and the last inequality is because the graph has no copies of $$K_{2,t}$$.

By using similar arguments, I got:

$$n \cdot\binom{\frac{2m}{n}}{t} \leq \sum_{v \in V} \binom{d(v)}{t}= \#K_{1,t} \leq \binom{n}{t}$$

But here I'm having some trouble getting to the final result. Is this the right way to go? any help would be appreciated!

It's a right way to go in the sense that it will give you a bound on $$m$$, but it will not give you the best bound. You've chosen the wrong generalization of $$K_{2,2}$$ to $$K_{2,t}$$. Instead of looking at $$t$$-element subsets of each neighborhood (which must cover each of $$\binom nt$$ $$t$$-element sets at most once) you should continue looking at $$2$$-element subsets of each neighborhood (which now cover each of $$\binom n2$$ $$2$$-element sets at most $$t-1$$ times).
Then, by using convexity in exactly the way you have been doing, we have $$n \binom{2m/n}{2} \le \sum_{v \in V} \binom{d(v)}{2} \le (t-1) \binom n2.$$ Or, to avoid dealing with inconvenient factors, we can write a slightly weaker but simpler inequality: $$n \cdot \frac{(2m/n - 1)^2}{2} \le (t-1) \cdot \frac{n^2}{2}.$$ Solving for $$m$$, you will get the bound you want.
There's nothing special about $$2$$, and in fact the same approach gives you the Kővári–Sós–Turán theorem: that an $$n$$-vertex graph with no copies of $$K_{s,t}$$ has at most $$O(n^{2-1/s})$$ edges. Just as in the case of $$K_{2,t}$$, there are two ways to look at $$K_{s,t}$$, and the other way gives a bound of $$O(n^{2-1/t})$$; we use whichever is better. However, for $$K_{2,t}$$, a result of Füredi gives a matching lower bound construction, which is not known to exist in general.