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?

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 :

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

