Let $k$ be an integer. The following inequality is standard. 
$$
(a+b)^{k+1} - b^{k+1} \leq (k+1)a(a+b)^k
$$
for $a,b > 0$.


However, does the following inequality still hold
$$
(a+b)^{k+1} - b^{k+1} \leq (k+1)a\left(a+ \frac{b}{(k+1)^{1/(k+1)}} \right)^k
$$
for $a,b > 0$? While $k \rightarrow \infty$, the term $(k+1)^{1/(k+1)} \rightarrow 1$ so that becomes the first inequality. What about if $k$ is large enough? 



Strengthening an inequality
Let's denote $x = \frac{b}{a}$, then the second inequality may be rewritten in following form
$$ (1+x)^{k+1} - x^{k+1} \leqslant (k+1) \left(1+ \frac{x}{(k+1)^{\frac{1}{k+1}}}\right)^{k}, $$
but the leading coefficient of RHS is equal to $(k+1)^{\frac{1}{k+1}}$, while in the LHS it is $k+1$, which is strictly greater unless $k=0$, so inequality fails for large values of $x$.


