Upper bound estimate $ab^3\leq c(a^2+b^2)^2$. Find optimal, or good, constants $c$

Question

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.)

Answer 1

Let $b=ka$ for some $k\in\Bbb R$. Then $$\frac{ab^3}{(a^2+b^2)^2}=\frac{k^3a^4}{(a^2+k^2a^2)^2}=\frac{k^3}{(1+k^2)^2}=\frac1{k^{-3}+2k^{-1}+k}$$ which attains its maximum when the denominator is minimised (derivative is zero). Thus $$-\frac3{k^4}-\frac2{k^2}+1=0\implies(k^2+1)(k^2-3)=0\implies k=\pm\sqrt3$$ so $c=\dfrac{3\sqrt3}{16}$.

Answer 2

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 $c=\frac{3\sqrt3}{16}.$

Answer 3

Another way: As this is homogeneous, we may set $a^2+b^2=1$, whence it is maximising $\sin t \cos^3t$, which is easy to see happens when say $t=\pi/6$.