1. ## inequalities

If$\displaystyle x+y = 3z$ prove that $\displaystyle x^2 + y^2 >= 3z^2$

Thanks

2. Note that $\displaystyle 3z^2=3(3z/3)^2$, so one has to show that $\displaystyle x^2+y^2\ge3((x+y)/3)^2$. After multiplying both sides by 3 and moving everything to the left, that expression can be represented as a sum of two squares.

3. Originally Posted by emakarov
Note that $\displaystyle 3z^2=3(3z/3)^2$, so one has to show that $\displaystyle x^2+y^2\ge3((x+y)/3)^2$. After multiplying both sides by 3 and moving everything to the left, that expression can be represented as a sum of two squares.
Maybe I have a blind spot, but I don't see that last step of getting to the sum of two squares. Is there some identity that I've forgotten ? .

I can get to the result using the AM >= GM inequality, (so $\displaystyle x^{2}+y^{2}\geq 2xy).$

4. \displaystyle \begin{aligned} x^2+y^2&\ge3((x+y)/3)^2\Leftrightarrow{}\\ 3x^2+3y^2-x^2-2xy-y^2&\ge0\Leftrightarrow{}\\ x^2-xy+y^2&\ge0\Leftrightarrow{}\\ (x-y/2)^2+3y^2/4&\ge0 \end{aligned}

5. Nice one. Thanks, I didn't see that .