Why is this flagged?

Is there a specific theorem to prove a^2 + b^2 > ab?

I will assume that a and be are in the set of real numbers.

If and a and b are both  0 then it is not true

If one is zero but not the other than LHS is positive and RHS is 0 so it will be true

If a or b (but not both) is negative then the RHS is positive and the LHS is negative so it will be true

So I am just left to prove it is true if a and b both have the same sign. (and not 0)

$(a-b)^2=a^2+b^2-2ab\\so\\ LHS=a^2+b^2\\ LHS=(a-b)^2+2ab\\ LHS=ab+[(a-b)^2+ab]\\ LHS=ab + \text{a positive number}\\ LHS >ab\\ LHS>RHS$

Yes, there is a theorem that states that$a^2 + b^2 > ab$ for any real numbers a and b. This theorem is known as the Cauchy-Schwarz inequality.

The Cauchy-Schwarz inequality states that for any two sequences of real numbers $(a_1, a_2, ..., a_n)$ and $(b_1, b_2, ..., b_n)$, we have:

$(a_1^2 + a_2^2 + ... + a_n^2)(b_1^2 + b_2^2 + ... + b_n^2) >= (a_1 b_1 + a_2 b_2 + ... + a_n b_n)^2$

Taking $n = 2$, we have:

$(a^2 + b^2)(b^2 + b^2) >= (ab + ab)^2$

Expanding the right-hand side and cancelling the $b^2$ terms, we get:

$a^2 + b^2 > ab$

So, for any real numbers a and b,$a^2 + b^2 > ab$.

It's a very useful and important theorem in mathematics and has many applications in various fields such as linear algebra, functional analysis, and optimization.