Problem 1032

Solution. We have,

$$\begin{aligned} & a^{5}+a^{4}+a^{3}+a^{2}+a+1=\left(a^{3}+1\right)\left(a^{2}+a+1\right) \text { and } \\ & b^{5}+b^{4}+b^{3}+b^{2}+b+1=\left(b^{3}+1\right)\left(b^{2}+b+1\right) \end{aligned}$$

Therefore, the given inequality is equivalent to the inequality

$$\left(a^{3}+1\right)\left(b^{3}+1\right) \geq 4$$

However, since $a, b$ are positive real numbers, from the inequality between the arithmetic and geometric means and the equality $a b=1$, it follows that

$$\left(a^{3}+1\right)\left(b^{3}+1\right) \geq 2 \sqrt{a^{3}} \cdot \sqrt{b^{3}}=4 \sqrt{(a b)^{3}}=4$$

Clearly, the equality holds if and only if $a=b=1$.

## 4th Year

1AB. If the solutions of the equation $x^{3}-p x^{2}+q x-r=0$ form a geometric progression, prove that $q^{3}=p^{3} r$.