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I'm interested in the following inequality : let $x\in [0,1]$ and $k\geq 1$, $$ (1+x)^k \leq 1+(2^k -1)x.$$ How to tackle this inequality ? Is it "obvious" ?

Thanks a lot.

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    Have you tried to prove it by induction? Another simpler way to prove it is using the fact that $(1+x)^k \le 1+kx$ and $k+1 \le 2^k$ (which are proved by induction as well). – 2017-01-05
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    @Crostul no i didn't try. Thank you. Though, the original inequality was for $k\geq 1$ real. – 2017-01-05
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    @Crostul Bernoulli's inequality assures $(1+x)^k\ge 1+kx$ for real $k\ge 1$. Besides induction won't work on its own for reals. – 2017-01-05

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Hint: For $k\ge1, x\ge0$, the function $f(x)=(1+x)^k$ is convex. In particular, successive secants have increasing slope, or the secant between endpoints of $[0, x]$ has lower slope than that between those of $[0,1]$.

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    P.S. Above method proves for all real $k\ge 1$... – 2017-01-05
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Induction does the trick. For $k = 0$ both sides equal $1$. No suppose, the equation holds for $k$, we then have \begin{align*} (1+x)^{k+1} &= (1+x)(1+x)^k \\ &\le (1+x)\bigl(1+(2^k-1)x\bigr)\\ &\le 1 + 2^kx + (2^k -1)x^2\\ &\le 1 + 2^kx + (2^k - 1)x\\ &= 1 + (2^{k+1} - 1)x \end{align*}

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    Thanks a lot, it was kind of obvious afterall. I also add that the original inequality was for $k\geq 1$ real. – 2017-01-05
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Since $x\in [0, 1]$, \begin{align*} (1 + x)^k &= \sum_{n=0}^k \binom{k}{n} x^n = 1 + \sum_{n=1}^k \binom{k}{n} x^n \leq 1 + x\sum_{n=1}^k \binom{k}{n} = 1 + (2^k - 1)x. \end{align*}