Suppose $K_{m,n}$ is a complete bipartite graph. Then $mn\ge m+n-1$.
Taking $m\ge1$ fixed. Applying induction on $n$.
$n=1$, Result hold trivially.
Assume result holds for $n=k$.
So, $mk \ge m+k-1$. We want to prove for $n=k+1$.That is we are adding one more vertex to other partite sets. So the number of edges will be increased by $m$. We know that $m\ge 1$. Adding $m$ on both sides of our assumed inequality. We get, $mk+m \ge m+k-1+m$. Also we have $m\ge 1$. Add $m+k-1$ on both sides. We get $m+k-1+m\ge m+k$. Hence, by induction, Result holds for all positive integers. $\because$ $m,n$ are symmetric, we can do this for $n\ge 1$. Did I write the correct proof for the given result? How do I prove the result without induction?
By König's theorem, if $\kappa_i,\lambda_i$ $(i\in I)$ are cardinal numbers such that $\lambda_i\gt\kappa_i$ for each $i\in I,$ then $\prod_{i\in I}\lambda_i\gt\sum_{i\in I}\kappa_i.$ In particular, if $a_1,a_2,b_1,b_2$ are nonnegative integers such that $a_1\gt b_1$ and $a_2\gt b_2,$ then $a_1b_1\gt a_2+b_2.$ If $m,n$ are positive integers then, since $m\gt m-1\ge0$ and $n\gt n-1\ge0,$ it follows that $mn\gt(m-1)+(n-1)=m+n-2.$ Since $mn$ is an integer, this means that $mn\ge m+n-1.$