1
$\begingroup$

We have $x, y, z \in \mathbb R$ such that $x+y+z=1$. Prove that $x^2+y^2+z^2 \geq \frac{1}{3}$. I am able to do this using the relationship between the power and arithmetic means. Is there a way to not use this relationship?

$\endgroup$
3
$\begingroup$

Hint: $(x-\frac13)^2 + (y-\frac13)^2 + (z-\frac13)^2 \geq 0$.

$\endgroup$
  • 1
    $\begingroup$ You could've edited your previous answer rather than deleting it and writing a new one. $\endgroup$ – Arthur Oct 6 '16 at 19:14
1
$\begingroup$

We have $$ x+y+z=1\\ (x+y+z)^2=1\\ x^2+y^2+z^2+2(xy+xz+yz)=1 $$ By the rearrangement inequality, $x^2+y^2+z^2\geq xy+xz+yz$. Inserting that gives you $$ 3(x^2+y^2+z^2)\geq1 $$

$\endgroup$
0
$\begingroup$

For any real numbers $x,y,z$, we have

$$\begin{align} 0\le(x-y)^2+(y-z)^2+(z-x)^2&\implies2(xy+yz+zx)\le2(x^2+y^2+z^2)\\ &\implies(x+y+z)^2\le3(x^2+y^2+z^2) \end{align}$$

So if $x+y+z=1$, then ${1\over3}\le x^2+y^2+z^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.