Proving a variable always returns true in an inequality I have the following inequality 
$$(1+a)^n\leq 1+(2^n−1)a$$ for $0 ≤ a ≤ 1$
What's the approach to prove that any value of the variable returns true ?
 A: We will prove this by induction on $n$, assuming $n > 0$. First, the base for the induction, $n=1$:
$$
(1 + a)^1 \leq 1 + (2^1 - 1)a
\iff 1 + a \leq 1 + a
$$
which is true. Now, the induction hypothesis is $(1 + a)^k \leq 1 + (2^k - 1)a$. We can add $2^ka$ to both sides to get
$$\begin{align}
1 + (2^{k+1} - 1)a &\geq (1 + a)^k + 2^ka \\
&\geq (1 + a)^k + (1 + a)^ka \\
&= (1 + a)^{k+1}
\end{align}$$
and we are done. Note that we used the fact that $a \leq 1$ when we wrote $2^k \geq (1 + a)^k$ and the fact that $0 \leq a$ when we multiplied this by $a$.
For negative $n$, the statement becomes $\frac{1}{(1 + a)^n} \leq 1 + \big(\frac{1}{2^n} - 1\big)a$. The proof proceeds almost exactly as before; for the base, at $n = 1$,
$$\begin{align}
\frac{1}{(1 + a)^1} &\leq 1 + \bigg(\frac{1}{2^1} - 1\bigg)a \\
\iff \frac{1}{1 + a} &\leq 1 - \frac{a}{2} = \frac{2 - a}{2} \\
\iff 2 &\leq -a^2 + a + 2 \\
\iff a &\leq 1
\end{align}$$
Assuming $\frac{1}{(1 + a)^k} \leq 1 + \big(\frac{1}{2^k} - 1\big)a$ and subtracting $\frac{a}{2^{k+1}}$ from both sides , we have
$$\begin{align}
1 + \big(\frac{1}{2^{k+1}} - 1\big)a &\geq \frac{1}{(1 + a)^k}-\frac{a}{2^{k+1}} \\
&\geq \frac{1}{(1 + a)^k}-\frac{a}{(1 + a)^{k+1}} \\
&= \frac{1}{(1 + a)^{k+1}}
\end{align}$$
as desired. We used $0 \leq a \leq 1$ in the base case as well as in the inductive step, so that should not be an issue. Note that the double inversion of a negative and a reciprocal allows us to use $2 \geq 1 + a$ as before.
This proves the result for $n \geq 1$ and $n \leq -1$. For $n = 0$, the result is trivial and boring, but completes the proof for all integers $n$.
A: In general, to prove statement $P(n)$ is true for all $n$ using induction principle, one start with a base case, say $n = 1$, prove $P(n)$ is true for $n = 1$, then assume $P(n)$ is true for $n = k$, based on that, proof $P(n)$ is true for $n = k+1$.  That completes the proof.  There are variations of the induction principle.  You might want to read a bit more on that.
For your case, $P(n)$ is $(1+a)^n\leq 1+(2^n−1)a$.  It is easy to verify $P(n)$ is true for $n = 1$.  Now assume $P(n)$ is true for $n = k$, that is $(1+a)^k \le 1 + (2^k-1)a$.  Now let's look at the case of $n = k+1$, since we assumed $P(k)$ is true, and $0 \le a \le 1$, 
$$(1+a)^{k+1} = (1+a)^k(1+a) \le \left(1+(2^k-1)a\right)(1+a).$$
We know that
$$\left(1+(2^k-1)a\right)(1+a) = 1+a + (2^k-1)a + (2^k-1)a^2 \le 1+a+(2^k-2)a \le 1+(2^{k+1}-1)a$$
Combine them, we have
$$(1+a)^{k+1} \le 1+(2^{k+1}-1)a,$$
which is exactly $P(n)$ for $n = k+1$.  QED.
A: There is equality for $n=1$ and any $a$. We take the logarithm on both sides. Then we treat $n$ as a continuous variable and show that the RHS grows faster with $n$ than the LHS for all $a$ and all $n \geq 2$, where $n$ does not have to be an integer. So we need to show 
$$
\frac{d}{ dn}\log ( (1+a)^n) \leq \frac{d}{ dn}\log (1+(2^n−1)a)
$$
$$
\log (1+a) \leq  \frac{a 2^n \log 2}{1+(2^n−1)a}
$$
With $
\log (1+a) < a $ for all $a$, it suffices to show 
$$
1\leq  \frac{ 2^n \log 2}{1+(2^n−1)a}
$$
or 
$$
1- a  \leq  { 2^n (\log 2 -a)}
$$
With $n\geq 2$, it is enough that 
$$
1- a  \leq  { 4(\log 2 -a)}
$$
which is fine for 
$a \leq (4\log 2 -1)/3\simeq 0.59$.
As another expansion, we use $
\log (1+a) = \log (2+a-1) = \log (2) + \log (1+(a-1)/2) \leq  \log (2) + (a-1)/2$ and get, as above, that it is enough to show
$$
\log (2) + (a-1)/2\leq  \frac{a  2^n \log 2}{1+(2^n−1)a}
$$
or
$$
(\log (2) + (a-1)/2 )({1-a})\leq {a  2^n (\log 2 - \log (2) - (a-1)/2 )}
$$
Again with $2^n \geq 4$, it suffices to show 
$$
(\log (2) + (a-1)/2 )({1-a})\leq {2 a  (1-a)}
$$
or
$$
2\log (2) -1  \leq {3 a }
$$
or
$$
a \geq (2\log (2) -1 )/3 \simeq 0.1288
$$
Combining  both approaches, the inequality is proved for all $n\geq 2$, where $n$ is not required to be integer.
$\quad \quad \Box$
