<s>[INST] 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?
 [/INST] 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).
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