## Task A-1.2.

Let $a$ and $b$ be positive real numbers such that

$$
\frac{a}{b}+\frac{b}{a}=3 \quad \text { and } \quad \frac{a^{2}}{b}+\frac{b^{2}}{a}=10
$$

Determine $\frac{1}{a}+\frac{1}{b}$.
## Solution.

By multiplying each of the equalities by $a b$ we have

$$
a^{2}+b^{2}=3 a b \quad \text { and } \quad a^{3}+b^{3}=10 a b
$$

Notice, $a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)$.

Substituting the first equality into the second, we get $10 a b=(a+b)(3 a b-a b)=2 a b(a+b)$, from which dividing by $2 a b$ yields $a+b=5$.

From $(a+b)^{2}=a^{2}+2 a b+b^{2}$ and the first equality, we have $(a+b)^{2}=5 a b$.

Substituting, we get $5 a b=5^{2}$, or $a b=5$.

By bringing to a common denominator and adding, we have $\frac{1}{a}+\frac{1}{b}=\frac{a+b}{a b}$.

Substituting the previously obtained values, we get $\frac{1}{a}+\frac{1}{b}=\frac{5}{5}=1$.

Note: The equality $a+b=5$ can be obtained in several ways. For example:

Multiplying the first equality by $a$ and by $b$ we have

$$
\frac{a^{2}}{b}+b=3 a \quad \text { and } \quad a+\frac{b^{2}}{a}=3 b
$$

Adding, we get $\frac{a^{2}}{b}+b+a+\frac{b^{2}}{a}=3(a+b)$.

Substituting the second equation, we have $a+b+10=3(a+b)$, or $a+b=5$.