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Using the AM and GM inequality, given that $a\gt0, b\gt0, c\gt0$ and $a+b+c=1$ prove that $$a^2+b^2+c^2\geqslant\frac{1}{3}$$

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$$a^2+{1\over 9} + b^2+{1\over 9} + c^2+{1\over 9}\geq {2\over 3}(a+b+c)$$ by AM-GM.

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  • $\begingroup$ I don't really understand how you obtained the result. Could you elaborate please? $\endgroup$ – Dr. Mathva Mar 30 at 19:16
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HINT: You can use your idea of squaring $a+b+c$, but also note that $\color{blue}{ab+bc+ca \le a^2 + b^2 + c^2}$, which you can prove with the help of AM-GM. (Hint for proving this: the AM-GM inequality tells us what about $a^2 + b^2, b^2+c^2$ and $c^2+a^2$?)

One more hint (based on a suggestion from user qsmy): let $x = a^2+b^2+c^2$ and $y = ab+bc+ca$. Squaring both sides of $a+b+c=1$ gives $x+2y=1$, and the blue inequality is $x\geq y$. Can you see it now?

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    $\begingroup$ I know the inequality that you stated, but I just can't seem to connect it with my question, please help. Thanks! $\endgroup$ – T. Joel Feb 17 at 2:16
  • $\begingroup$ If you expand $1=(a+b+c)^2$, you should find $ab+bc+ca$ pop up. Apply the blue inequality above to this term. $\endgroup$ – Minus One-Twelfth Feb 17 at 2:24
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In the worst case possible you'd get $$a = b = c = \frac{1}{3} \Longrightarrow a^2 + b^2 + c^2 = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} \geq \frac{1}{3} $$

In the best case possible you'd get $$a = 1, b = c = 0 \Longrightarrow 1^2 + 0^2 + 0^2 = 1 \geq 1/3 $$

Therefore the inequality holds. Didn't use the AM-GM inequality, though.

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