If $\displaystyle a^2+b^2=2abc$, prove that $\displaystyle |c|\geq a$ and $\displaystyle |c|\geq b$
I run out of ideas.
I tried $\displaystyle a^2+b^2+2ab=2ab+2abc $
I tried $\displaystyle a^2=2abc-b^2 $
Does anyone know how to prove this?
If $\displaystyle a^2+b^2=2abc$, prove that $\displaystyle |c|\geq a$ and $\displaystyle |c|\geq b$
I run out of ideas.
I tried $\displaystyle a^2+b^2+2ab=2ab+2abc $
I tried $\displaystyle a^2=2abc-b^2 $
Does anyone know how to prove this?
That result is not true. The result should be that $\displaystyle |c|\geqslant1$. For example, if a=3 and b=4 then c = 25/24, which is greater than 1, but less than a or b. Edit: HallsofIvy beat me to it!
To see why $\displaystyle |c|\geqslant1$, divide the equation by $\displaystyle b^2$ (assuming that $\displaystyle b\ne0$). It becomes $\displaystyle \bigl(\tfrac ab\bigr)^2 + 1 = 2\bigl(\tfrac ab\bigr)c$. Now put $\displaystyle x = \tfrac ab$, so that $\displaystyle x^2-2cx+1=0$. The condition for that quadratic to have real roots is $\displaystyle |c|\geqslant1$.