Find the biggest number $M$ such that the inequality $a^2+b^{1389} \ge Mab$ holds for every $a,b \in [0,1]$. Find the biggest number $M$ such that the following inequality holds for every $a,b \in [0,1]$,
$$a^2+b^{1389} \ge Mab$$
My attempt:We should find the minimum of $\frac{a}{b}+\frac{b^{1388}}{a}$.By putting $a=b=0^{+}$ we will get to $1 \ge M$ and clearly $M \ge 0$ but how much is $M$?

By AM-GM $\frac{a}{b}+\frac{b^{1388}}{a}\geq2b^{693.5}\rightarrow0^+$ for $b\rightarrow0^+$.
The equality occurs for $a=b^{694.5}$ and we see that $\{a,b\}\subset[0,1]$.
Thus, the answer is $0$.
