How to show that deleting at most $(m-s)(n-t)/s$ edges from a $K_{m,n}$ will never destroy all its $K_{s,t}$ subgraphs. This problem is from Graph Theory by Diestel Chapter 7 (Extremal Graph Theory) section 5 (regularity lemma) problem 9. 
I was thinking about applying the Erdos & Stone theorem, but I am honestly lost and I am not sure how to go about this problem. Any help will be greatly appreciated. 
 A: 
The Zarankiewicz function $z(m, n; s, t)$ denotes the maximum possible number of edges in a bipartite graph $G = (U, V, E)$ for which $|U| = m$ and $|V| = n$, but which does not contain a subgraph of the form $Ks,t$.

Thus it suffices to show that $z(m, n; s, t)\le mn-(m-s)(n-t)/s-1$. For this we have such a powerful tool as Kővári–Sós–Turán theorem.  By it, 
$$z(m, n; s, t)<(s-1)^{1/t}(n-t+1)m^{1-1/t}+(t-1)m.$$
So it suffices to show that 
$(s-1)^{1/t}(n-t+1)m^{1-1/t}+(t-1)m\le mn-(m-s)(n-t)/s-1$
$(s-1)^{1/t}(n-t+1)m^{-1/t}+t-1\le n-(1-s/m)(n-t)/s-1$
$\left(\frac{s-1}m\right)^t(n-t-1)+\left(\frac 1s-\frac 1m\right)(n-t)+t\le n$.
So it suffices to show that 
$\left(\frac{s-1}m\right)^t+\frac 1s-\frac 1m\le 1$.
Assuming $t,s\ge 1$ and $s-1\le m$ it suffices to show that 
$\frac{s-1}m+\frac 1s-\frac 1m\le 1$
$s^2-2s+m\le sm$
$s^2\le (s-1)m+2s$.
But $(s-1)m+2s\ge (s-1)^2+2s=s^2+1>s^2$.
PS. If one prefers a homemade proof to a civilized that, then we can consider the following arguments. Let the bipartition of vertices of the graph $K_{m,n}$ is constituted of $m$ red vertices $\{v_1,\dots, v_m\}$ and $n$ blue vertices. Assume that we deleted $d$ edges from the graph $K_{m,n}$ and the resulting graph $G$ contains no $K_{s,t}$ subgraphs. Let $1\le s\le m$, $1\le t\le n$, and $$\deg v_1\ge \deg v_2\ge \dots \ge \deg v_s\ge\dots \ge \deg v_m$$ be the degrees of the red vertices of $G$. Since the set of blue vertices contain no subset of size $t$ with each vertex adjacent to each of the vertices $\{v_1,\dots, v_s\}$ then 
$$(n-\deg v_1)+ (n-\deg v_2)+\dots+(n-\deg v_s)\ge n-t+1$$ 
$$ns-\sum_{i=1}^s \deg v_i\ge n-t+1$$ 
$$ns-n+t-1 \ge \sum_{i=1}^s \deg v_i$$ 
On the other hand, since $\deg v_1\ge \deg v_2\ge \dots \deg v_m$, 
$$\sum_{i=1}^s \deg v_i\ge\frac sm\left(\sum_{i=1}^m \deg v_i\right)= \frac sm|E(G)|= \frac sm (mn-d).$$   
$$ns-n+t-1\ge \frac sm (mn-d)$$
$$d \ge \frac{m(n-t+1)}{s}$$
So it suffices to show that 
$\frac{m(n-t+1)}{s}\ge \frac{(m-s)(n-t)}{s}+1$ 
$m(n-t+1) \ge (m-s)(n-t) +s$
$mn-mt+m \ge mn-mt-ns+st+s$ 
$ns+m \ge st+s$
which  follows from $n\ge t$ and $m\ge s$.
