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?
On
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 $xyz > abc$.
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 $9+2+2 \gt 4+4+4, 36 \lt 64$