Young’s inequality implies $a^ {n/(n+1)}\leq 2a+(1/n^2)$ for any $a>0$ and $n\geq 1$

Question

I’ve seen this statement in a Yano’s article and I can not prove it. I take the Young’s inequality $a^{1/p}b^{1/q} \leq a/p + b/q$ where $1/p +1/q =1$. I’ve prove it in the case $a\geq 1$. In the case where $n=1$ you can deduce it inmediatly from the Young’s inequality but in other cases I can not see the proof.

Answer 1

The idea is to find a “suitable” real number $c > 0$ such that the application of Young's inequality to $$ a^{n/(n+1)} = (ca)^{n/(n+1)} \, \cdot \, (1/c^n)^{1/(n+1)} \le \frac{n}{n+1} ca + \frac{1}{n+1} \frac{1}{c^n} $$ gives the term $2a$ as the first summand on the right-hand side.

This is the case for $c=\frac{2(n+1)}{n}$, so that we have $$ \tag{*} a^{n/(n+1)} \le 2a + \frac{1}{n+1}\left( \frac{n}{2(n+1)}\right)^n \, . $$ That is stronger than the desired estimate because $$ \frac{1}{n+1}\left( \frac{n}{2(n+1)}\right)^n < \frac{1}{(n +1) 2^n} < \frac{1}{(n+1)^2} < \frac{1}{n^2} \, . $$

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An alternative is to observe that for fixed $n$ the function $f(a) = a^{n/(n+1)}$ is concave, so that we have $$ f(a) \le f(b) + (a-b)f'(b) \quad \text{for all } a, b > 0 \, . $$ Choosing $b$ such that $f'(b) = 2$ gives again the inequality $(*)$.