<s>[INST] If $r = \min \left\{ a , \frac{b}{ a+ a^2 b^3} \right\}$ , find $ r_{\max}$ Studying differential equations I came cross through this:
Let $ \displaystyle{ r = \min \left\{ a , \frac{b}{ a+ a^2 b^3} \right\} } $, where $ a,b >0$. Find $ r_{ \max} $.
Here is what I did:
Fix $ a> 0$ and define $ \displaystyle{ g(b) = \frac{b}{ a+ a^2 b^3} ,\quad b>0 }$. Then we have that $ \displaystyle{ g'(b) = \frac{ a(1-2ab^3)}{ (a+ a^2 b^3)^2} }$.
Thus,  $g$ has maximum at $ \displaystyle{b_0= \frac{1}{ \sqrt[3]{2a} }}$ which gives $ \displaystyle{ g(b) \leq g\left( \frac{1}{ \sqrt[3]{2a}} \right) = \frac{2}{ 3a \sqrt[3]{2a} }}$
So, $ \displaystyle{ r = \min \{ a , \frac{2}{ 3a \sqrt[3]{2a} } \}}$. I think that to find $a$ which gives $ r_{\max}$ I have to solve the equation $ \displaystyle{ a=\frac{2}{ 3a \sqrt[3]{2a}} }$.
Is this correct and why....?
Any ideas?
Thank's in advance!
edit: I edit the title.
 [/INST] It looks correct to me. The reason that is correct is because you're finding when the two values are equal. If they aren't equal, then since the real numbers are totally ordered, we know that one will be less than the other.
Solving for $a$, we have $$a^{7/3}=\frac{2^{2/3}}{3}.$$ Taking the $3/7$ root of each side yields $$a=\left(\frac{2^{2/3}}{3}\right)^{3/7}=\left(\frac{4}{27}\right)^{1/7}.$$
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