If $a,b,c$ are distinct positive real numbers and $a^2+b^2+c^2=1$, then prove $ab+bc + ac$ will be less than 1.

Question

$$(a+b+c)^2-2(ab+bc +ac)=1$$ $$ab+bc + ac=\frac{(a+b+c)^2-1}{2}$$ Which shows that it can take any positive real value. Is there anything else that needs to be added to get the answer?

Answer 1

I believe your question is "what is wrong with my answer" rather than "how do I get the correct answer" which I think you already know.

In one sense, there is nothing wrong with your answer of $$ab+bc + ac=\frac{(a+b+c)^2-1}{2}.$$ You have correctly expressed the variable quantity $ab+bc+ca$ in terms of another variable quantity.

The problem is that $\frac{(a+b+c)^2-1}{2}$ is dependent on the values of $a,b$ and $c$ which satisfy $a^2+b^2+c^2=1$ in just as complicated a way as $ab+bc+ca$.

So the idea of this type of question is to express quantities in terms of particular quantities (dependng on what the question is.) For this problem the best quantity would seem to be $(a-b)^2+(b-c)^2+(c-a)^2$.

Answer 2

Hint:

$$(a-b)^2+(b-c)^2+(c-a)^2>0$$

Answer 3

By the Rearrangement inequality: $$ab+bc+ca\le a^2+b^2+c^2=1.$$ If $a,b,c$ are mutually distinct, then the inequality will be strict.