Find the supremum of the set $ \lbrace abcd - \frac{a^2}{2}-\frac{b^{4}}{4}-\frac{c^8}{8}-\frac{d^{16}}{16} \rbrace $

Question

Find the supremum of the expression $$ abcd - \frac{a^2}{2}-\frac{b^{4}}{4}-\frac{c^8}{8}-\frac{d^{16}}{16} $$

if $a,b,c,d\in \mathbb{R}$

Answer 1

Apply AM $\ge$ GM to 8 copies of $a^2$, 4 copies of $b^4$, 2 copies of $c^8$, 1 copy of $d^{16}$ and 1 copy of $1$, we obtain

$$\frac{a^2}{2} + \frac{b^4}{4} + \frac{c^8}{8} + \frac{d^{16}}{16} + \frac{1}{16} \ge ((a^2)^8 (b^4)^4 (c^8)^2 (d^{16}))^{1/16} = |abcd| $$ As a result, $$abcd - \left(\frac{a^2}{2} + \frac{b^4}{4} + \frac{c^8}{4} + \frac{d^8}{16}\right) \le |abcd| - \left(\frac{a^2}{2} + \frac{b^4}{4} + \frac{c^8}{4} + \frac{d^8}{16}\right) \le \frac{1}{16}$$ Since the value $\frac{1}{16}$ is achieved at $a = b = c = d = 1$, the maximum of the set is $\frac{1}{16}$.