Bounding a simple expression

Question

Hi does anyone know what constant $C$ I could put here to make this inequality correct : for $a\in [0,1] $ $$ a^{1/5}+a^{4/21} \leq Ca^{4/21} $$

Answer 1

(Under the assumption $0\leq a\leq 1$) Rearranging, we find $$ a^{1/105}\leq C-1, $$ so we clearly need $C\geq 1$. On the other hand, the left hand side is increasing and assumes its maximum in $a=1$, hence $C=2$ is okay.

Answer 2

(This answers the original version of the question. For the current version, see the alternative answer.)

This is impossible.

• For any $C > 1$, $$a^{1/5}+a^{4/21} \leq Ca^{4/21} \iff a^{1/105} \le a^{1/5-4/21} \leq C-1,$$ so any $a > (C-1)^{105}$ violates the above inequality.
• For any $C \le 1$, $$a^{1/5}+a^{4/21} \leq Ca^{4/21} \le a^{4/21} \iff a^{1/5} \le 0,$$ so any $a > 0$ violates the above inequality.