Find the biggest number $M$ such that the inequality $a^2+b^{1389} \ge Mab$ holds for every $a,b \in [0,1]$.

Question

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 $M$?

Answer 1

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 $0$.

Answer 2

The condition is equivalent to $a^2 - M ab + b^{1389} \ge 0$ for $a,b \in [0,1]$. Considering it as a quadratic in $a$ and noting that $a=b=1$ gives $M \le 2\,$, its minimum is attained at $\frac{Mb}{2} \in [0,1]$ so in order for the inequality to hold the quadratic must have no distinct real roots i.e. $\Delta = M^2b^2 - 4 b^{1389} \le 0$. For $b \ne 0$ the latter gives $M^2 \le 4 b^{1387}$ and with $b \to 0^+$ it follows that $M^2 \le 0$ therefore $M=0$.