<s>[INST] Upper bound estimate $ab^3\leq c(a^2+b^2)^2$. Find optimal, or good, constants $c$ I am trying to find a good upper bound estimate for the expression $ab^3$, where $a,b\in\mathbb R$, and it should be of the form $ab^3\leq c (a^2+b^2)^2, c\in\mathbb R.$
(The reason for the latter is, because in my application of this, $a^2+b^2$ is a physical quantity. So I would like to express the estimate in terms of it.)
By trial and error experiments with Maple, I found that $c=1/3$ yields quite a good estimate. But when I try to derive an estimate analytically, the best I got so far was $c=1/2$, since $ab\leq\frac{1}{2}(a^2+b^2)$ gives
$$b^2\leq a^2+b^2\implies ab^3=(ab)b^2\leq\frac{1}{2}(a^2+b^2)^2.$$
How can I either get the better $c=1/3$ estimate analytically, or prove it? (Or an even better one would do as well of course.) 
 [/INST] By AM-GM
$$(a^2+b^2)^2=\left(a^2+3\cdot\frac{b^2}{3}\right)^2\geq\left(4\sqrt[4]{a^2\left(\frac{b^2}{3}\right)^3}\right)^2=\frac{16}{3\sqrt3}|ab^3|\geq\frac{16}{3\sqrt3}ab^3.$$
The equality occurs for $a=\frac{b}{\sqrt3},$ which says that $c=\frac{3\sqrt3}{16}.$
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