# Thread: Inequality involving absolute value

1. ## Inequality involving absolute value

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?

2. Originally Posted by novice
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?
It's good that you can't prove this because it is not true! Take a= b= 3 and c= 1. Then $\displaystyle a^2+ b^2= 9+9 18= 2(3)(3)(1)= 2abc$ but |c|= 1< a= 3 and |c|= 1< b= 3.

3. Originally Posted by novice
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 $\displaystyle |c|\geqslant1$.

4. Thank you both, HallsofIvy and Opalg. It's very nice to have different ways of seeing things. I sure have a lot to learn still.

5. If a^2 + b^2 = R^2,
2abc = R^2