Well from our assumption of the k-variable case, we have as a FACT that the statement is true for ANY k arbitrary non-negative real numbers x_1, x_2, ..., x_k. So if it's true for ANY of them, then it would certainly be true for k specific ones. I will only pick one of them to be specified, and will choose x_k = AM of the other k-1 terms. It all stems from the assumption that it is true for ANY k-ARBITRARY non negative real numbers.
@@MathemaEducation I'm sorry for the inconvenience, but i still don't get it, i'm not that good at maths and i'm trying to hopefully get better, is there any chance you could please re-explain it for me?
Waaaaaaaaao what a nice proof.
Love this proof!
i didnt get the logic behind assuming Xk =the AM of Xk-1 how could you be sure that it will be equal to 1
Well from our assumption of the k-variable case, we have as a FACT that the statement is true for ANY k arbitrary non-negative real numbers x_1, x_2, ..., x_k. So if it's true for ANY of them, then it would certainly be true for k specific ones. I will only pick one of them to be specified, and will choose x_k = AM of the other k-1 terms.
It all stems from the assumption that it is true for ANY k-ARBITRARY non negative real numbers.
@@MathemaEducation I'm sorry for the inconvenience, but i still don't get it, i'm not that good at maths and i'm trying to hopefully get better, is there any chance you could please re-explain it for me?
🤩