Prove an inequality involving powers of two Let $a,b,c$ be mutually distinct positive integers.
Prove 
$$2^{a+b} + 2^{b+c} + 2^{a+c} < 2^{a+b+c} + 1$$
Any hints on how to start proving it?
 A: I assume the set of Natural Numbers to be $\mathbb N = \{1,2,3 \ldots \}$
Let $$\alpha =2^{-a} + 2^{-b} + 2^{-c}$$
And 
$$\beta =1+2^{-(a+b+c)}$$
We can easily see that the maximum value of $\alpha$ will occur for minimum values of $a,b$, and $c$.
The minimum values of $a,b$ and $c$ , are $1,2$, and $3$ (All are distinct).
Hence $$\alpha =2^{-a} + 2^{-b} + 2^{-c} \le 2^{-1} + 2^{-2} + 2^{-3}=\frac 78$$
And we also have $\beta$ :
$$2^{-(a+b+c)} \gt 0 \implies \beta= 1+2^{-(a+b+c)} \gt 1$$
Thus we have :

$$\alpha \le \frac78 ~~{and}~~ \beta \gt 1$$

Hence, obviously $\alpha \lt \beta$
Now multiplying both the sides by any positive quantity won't change the inequality, so we multiply both the sides by $2^{a+b+c}$
Hence, finally we get :

$$2^{a+b} + 2^{b+c} + 2^{a+c} < 2^{a+b+c} + 1$$

A: Note that the left-hand side is a sum of distinct powers of $2$, so (assuming WLOG that $a < b < c$)
$$2^{a+b}+2^{b+c}+2^{a+c}<\sum_{k=0}^{b+c}2^k=2^{b+c+1}-1<2^{a+b+c}+1.$$
A: It can also be proven by induction.
Assume it's true for $(a,b,c)$. Then:
\begin{align}
2^{a+b+(c+1)}+1&=2\cdot 2^{a+b+c}+2-1\\
&=2\cdot 2^{a+b+c}+2-1\\
&=2(2^{a+b+c}+1)-1\\
&>2(2^{a+b}+2^{b+c}+2^{c+a})-1\\
&>2^{a+b}+2^{b+(c+1)}+2^{(c+1)+a}+2^{a+b}-1\\
&>2^{a+b}+2^{b+(c+1)}+2^{(c+1)+a}\\
\end{align}
Note that we didn't assume $a,b,c$ are distinct, or anything about the order. Thus, from $(a,b,c)$ we can not only derive $(a,b,c+1)$, but also $(a,b+1,c)$ and $(a+1,b,c)$. Now check the base case:
$$2^{1+2}+2^{2+3}+2^{3+1}<2^{1+2+3}+1$$
so that it's true for all $a,b,c\in\Bbb N$ (such that they are larger than $1$, $2$ and $3$; they don't have to be distinct, but we still needed that for the base case).
