<s>[INST] 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 ?
 [/INST] 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 $n$.
</s>