Minimum Value Optimization 3
Jonathon is given the following problem: Let $ a $ and $ b $ be positive real numbers such that $ a + b = 1 $. Find the minimum value of
\[\left( a + \frac{1}{a} \right)^2 + \left( b + \frac{1}{b} \right)^2.\]Jonathon reasons as follows: By AM-GM,
\[a + \frac{1}{a} \ge 2,\]and
\[b + \frac{1}{b} \ge 2,\]so
\[\left( a + \frac{1}{a} \right)^2 + \left( b + \frac{1}{b} \right)^2 \ge 2^2 + 2^2 = 8.\]He concludes that the minimum value is 8, but this answer is incorrect. What is the minimum value?
As a bonus, determine where Jonathon went wrong.
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Solution
By QM-AM,
\[\sqrt{\frac{\left( a + \frac{1}{a} \right)^2 + \left( b + \frac{1}{b} \right)^2}{2}} \ge \frac{\left( a + \frac{1}{a} \right) + \left( b + \frac{1}{b} \right)}{2} = \frac{\frac{1}{a} + \frac{1}{b} + 1}{2}.\]By AM-HM,
\[\frac{a + b}{2} \ge \frac{2}{\frac{1}{a} + \frac{1}{b}},\]so
\[\frac{1}{a} + \frac{1}{b} \ge \frac{4}{a + b} = 4.\]Then
\[\sqrt{\frac{\left( a + \frac{1}{a} \right)^2 + \left( b + \frac{1}{b} \right)^2}{2}} \ge \frac{5}{2},\]so
\[\left( a + \frac{1}{a} \right)^2 + \left( b + \frac{1}{b} \right)^2 \ge 2 \left( \frac{5}{2} \right)^2 = \frac{25}{2}.\]Equality occurs when $ a = b = \frac{1}{2}, $ so the minimum value is $ \boxed{\frac{25}{2}} $.
The reason that Jonathon's solution doesn't work is because
\[a + \frac{1}{a} = 2\]only when $ a = 1, $ and similarly
\[b + \frac{1}{b} = 2\]only when $ b = 1 $. Since $ a + b = 1, $ both conditions cannot hold simultaneously, which means that the expression in the problem cannot actually attain the value of 8. Jonathon's reasoning only shows that expression must be greater than or equal to 8, which is not enough to establish its minimum value.
This is why it is important to establish that the minimum/maximum value that you come up with can actually be attained. Here, we made sure to state that equality occurs when $ a = b = \frac{1}{2} $. So checking that the minimum/maximum value that you come up can be attained is not just a formality.
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