Integer inequality: $x + y +z> a + b + c$ does not imply $xyz > abc$ Prove by contradiction that for any integers $x,y,z,a,b,c$ greater than $0$ such that $x+y>a+b$, it is not implied that $x\cdot y\cdot z>a\cdot b\cdot c$?
Obviously this statement is true. Consider $9+1+1>3+2+2; 9<12$. But I cannot seem to prove it without falling back to examples. Also, would the outcome of the statement be different if I required that all integers be greater than $1$? If so, how do i go about formalizing that thought?  
 A: One example is sufficient to show that the implication is not true.  You probably mean $x + y + z \gt a+b+c \not \implies xyz \gt abc$.  For an example with all numbers greater than $1$, you again want a large disparity on the left and small one on the right.  For example $9+2+2 \gt 4+4+4, 36 \lt 64$
A: Examples can be quite healthy, guarding against excessive abstraction. But in some cases, and this may or may not be one of them, what's needed is a way to create infinitely many examples. How about this? [various failed attempts omitted here]
EDIT: Okay, I think I've got it now. Choose a number $n$ with at least four distinct prime factors. Now choose six nontrivial divisors (neither 1 nor the number itself) of that number and label them $x, y, z, a, b, c$ such that $x + y + z > a + b + c$ yet $xyz = abc$. Now reassign $x := x - 1$, so now $x \geq 1$ and therefore the inequality $x + y + z > a + b + c$ still holds but now $xyz < abc$. A similar adjustment can be made so as to obtain $xyz > abc$.
