Suppose $a$ and $b$ are positive real numbers with $a > b$ and $ab = 8.$ Find the minimum value of $\frac{a^2 + b^2}{a - b}.$

Question

How would we go about minimizing this? We can plug in $b = 8/a$ and go about minimizing the new expression but it isn't clear how to minimize.

What do you do?

score 3 · Accepted Answer · answered Jun 17 '21 at 16:14

3

HINT

Since $a > b$, it results that $a - b > 0$. Consequently, we have that

\begin{align*} \frac{a^{2}+b^{2}}{a-b} & = \frac{(a-b)^{2} + 2ab}{a-b}\\\\ & = a - b + \frac{2ab}{a-b}\\\\ & = a - b + \frac{16}{a-b} \end{align*}

Now you can apply the AM-GM inequality.

Can you take it from here?

answered Jun 17 '21 at 16:14

user0102

Got it thank you :) – Patrick Jun 17 '21 at 16:58
@Patrick If this answered your question, mark the "answered" check next to it (and perhaps upvote it as well) – Alan Jun 17 '21 at 17:12

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