# Thread: Inequality Proof question

1. ## Inequality Proof question

Hi my first post! Feel embarrassed I cant see the solution as I though I was getting quite good at these.... Any here is the question:

If a2 + b2 = 1 = c2 + d2 ; a, b, c, d are elements of R

Prove that ab + cd < 1

Apologies I cant seem to use superscript; the number 2 in the question is meant to be to the power of 2 for each of the letters in the equation.

2. Originally Posted by Abromavich
Apologies I cant seem to use superscript; the number 2 in the question is meant to be to the power of 2 for each of the letters in the equation.
Why not learn to post in symbols? You can use LaTeX tags
$$x^{-2}+y^{10}$$ gives $\displaystyle x^{-2}+y^{10}$

$$ab+cd<1$$ gives $\displaystyle ab+cd<1$

Originally Posted by Abromavich
If $\displaystyle a^2 + b^2 = 1 = c^2 + d^2$ ; a, b, c, d are elements of R
Prove that $\displaystyle ab + cd < 1$
Note that $\displaystyle a^2+b^2\ge 2ab~\&~c^2+d^2\ge 2cd.$

3. Thanks will use latex from now on. How do you know that $\displaystyle ab=cd$

4. Originally Posted by Abromavich
Thanks will use latex from now on. How do you know that $\displaystyle ab=cd$
More importantly, we know that $\displaystyle 2ab\le a^2+b^2=1$.
Now add the two and divide by 2.

5. lol so simple, thanks for the help. Ill prob be back very soon