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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?

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up vote 2 down vote accepted

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 $x$.

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That is, for any given $k$ the inequality fails for large enough $x$. On the other hand, for given $x > 0$, the inequality is true for large enough $k$. –  Robert Israel Jul 19 '12 at 18:15
    
It is, indeed. What is also true, is the following estimate $$ (1+x)^{k+1} - x^{k+1} \leqslant (k+1)\left( (k+1)^{-\frac{1}{k+1}} + x\right)^k,$$ which follows by easy induction (we differentiate) relying on the fact that $n^{\frac{1}{n}}$ is decreasing for $n\geqslant 3$; we also have to examine the case $k=2$. –  Mateusz Wasilewski Jul 19 '12 at 19:02
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