Re: First Proofs class - HELP!

Quote:

Originally Posted by

**resjsu** Let x,y,z be nonnegative real numbers such that y+z > or = to 2.

Prove: (x+y+z)^2 is > or = 4x + 4yz

To do this you must know this:

$\displaystyle \left( {\forall a\;\& \;b} \right)\left[ {a^2 + b^2 \ge 2ab} \right]$.

$\displaystyle \begin{align*}(x+y+z)^2&=x^2+y^2+z^2+2xy+2xz+2yz\\ &\ge x^2+2x(y+z)+4yz \end{align*}$

Re: First Proofs class - HELP!

Hello,

Thank you replying.

Can you explain how you got the right side?

Where did the x^2 and the 2x(y+z) come from?

I understand that if y+z equals at least 2 then the term turns into the original 4x but how did you know to look at that. Thats the trouble im having I don't know where to look for clues. I would've never thought of that.

Re: First Proofs class - HELP!

Can someone please explain?

Re: First Proofs class - HELP!

Quote:

Originally Posted by

**resjsu** Can someone please explain?

Can YOU expand $\displaystyle (x+y+z)^2~?$

If not then you have no business even attempting this problem.

Re: First Proofs class - HELP!

Of course I can, I stated that in the first post, its the right side I am not following.