# Prove that $a^2+b^2+c^2\geqslant\frac{1}{3}$ given that $a\gt0, b\gt0, c\gt0$ and $a+b+c=1$, using existing AM GM inequality [closed]

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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• What have you tried? – Thomas Shelby Feb 17 at 1:42
• Using (a+b+c)^2 = 1 but I got stuck – T. Joel Feb 17 at 1:45
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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.

• I don't really understand how you obtained the result. Could you elaborate please? – Dr. Mathva Mar 30 at 19:16

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?

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

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.