4. Determine the minimum value of the expression $E(a, b)=a+b+\frac{1}{a b}$, where $a, b>0$ and $a+b \leq 1$.

Romeo Ilie
## Solution.

Method 1. $E\left(\frac{1}{2}, \frac{1}{2}\right)=5$. (1p)

$$
\begin{aligned}
& a+b+\frac{1}{a b}=(a+b)+\frac{1}{4 a b}+\frac{1}{4 a b}+\frac{1}{4 a b}+\frac{1}{4 a b}(\mathbf{2 p}) \\
& \geq 5 \cdot \sqrt[5]{\frac{a+b}{4^{4}(a b)^{4}}}(\mathbf{1} \mathbf{p}) \\
& \geq 5 \cdot \sqrt[5]{\frac{(a+b)^{8}}{4^{4}(a b)^{4}}}=5 \cdot \sqrt[5]{\left[\frac{(a+b)^{2}}{4(a b)}\right]^{4}} \\
& \geq 5 .(1 p)
\end{aligned}
$$

It follows that $\min _{a, b>0, a+b \leq 1} E(a, b)=5 .(1 \mathbf{p})$

Method 2. $E\left(\frac{1}{2}, \frac{1}{2}\right)=5$. (1p)

Let $a, b>0$, with $a+b \leq 1$. Denote $S=a+b \in(0,1]$. Since $\left.00, a+b \leq 1} E(a, b)=5 .(\mathbf{1} \mathbf{p})$