# 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 $$a^2+b^2+c^2\geqslant\frac{1}{3}$$

• What have you tried?
– cqfd
Commented Feb 17, 2019 at 1:42
• Using (a+b+c)^2 = 1 but I got stuck Commented Feb 17, 2019 at 1:45
• Where exactly did you get stuck with that attempt? What stopped you from progressing? And also, please edit your question post with this information as that makes it easier for new readers to catch up (they won't have to sift through comments). Commented Feb 17, 2019 at 1:50
• Commented Feb 17, 2019 at 8:31
• @Théophile: More Information here: How to search on this site? and here: Announcing a third-party search engine for Math StackExchange. Commented Feb 17, 2019 at 17:43

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! Commented Feb 17, 2019 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. Commented Feb 17, 2019 at 2:24

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

• I don't really understand how you obtained the result. Could you elaborate please? Commented Mar 30, 2019 at 19:16

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.