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?
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See also: https://math.stackexchange.com/questions/2247973/inequality-proof-is-there-a-simpler-way-to-solve-it – Martin Sleziak Apr 23 '17 at 21:29
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Hint: $(x-\frac13)^2 + (y-\frac13)^2 + (z-\frac13)^2 \geq 0$.
user133281
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1You could've edited your previous answer rather than deleting it and writing a new one. – Arthur Oct 06 '16 at 19:14
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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 $$ 3(x^2+y^2+z^2)\geq1 $$
Arthur
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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 ${1\over3}\le x^2+y^2+z^2$.
Barry Cipra
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