Prove that if real numbers a,b,c satisfy the following inequalities, they are positive??
2) ab+ac+bc>0
3) abc>0
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a) All three are negative, but then condition 1 will not match
b) Only one of the numbers is negative, but then condition 3 will not match
So the only thing left is for two of the numbers to be negative. So let’s assume a and b are negative, then for condition1 to be true, we conclude that c > abs(a+b). Now looking at condition 2:
ab + ac + bc > 0
ab + c(a + b) > 0
In order for this condition to be true, then ab (a positive number) must be greater than abs(c(a+b)), and since c is positive: c*abs(a+b). But we know that c > abs(a+b) therefore:
c * abs(a + b) > abs(a + b) * abs(a + b) = a^2 + b^2 + 2ab > ab.
A contradicton which will make cndition 2 false.
Therefore the only possibility is for a, b and c to be positive.
