This question was uploaded on 17/12/21 on social media accounts.
This question is from Singapore Math Olympiad 2019 Open Round 1 (Question 11)
Solution:
Given equation:
\[1000a + \frac1{1000(a-b)b}\]
\[= 1000a\color{red} {-1000b +1000b} + \frac1{1000(a-b)b}\]
\[= 1000(a-b) + 1000b + \frac1{1000(a-b)b}\]
Now we have 3 terms and also we know that:
\[AM\geq GM\]
Let us consider these three terms:
\[1000(a-b),⠀1000b,⠀\frac1{1000(a-b)b}\]
\[\Rightarrow \frac {1000(a-b) + 1000b + \frac1{1000(a-b)b}}{3} \]
\[\geq \sqrt[3]{1000(a-b)\times1000b\times\frac1{1000(a-b)b}} = \sqrt[3]{1000}\]
\[\Rightarrow \frac {1000(a-b) + 1000b + \frac1{1000(a-b)b}}{3} \geq 10\]
\[\Rightarrow 1000(a-b) + 1000b + \frac1{1000(a-b)b} \geq 30\]
\[\Rightarrow 1000a + \frac1{1000(a-b)b} \geq 30\]
It means the minimum value is 30.
