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

## closed as off-topic by Eevee Trainer, Théophile, Martin R, José Carlos Santos, Kemono ChenFeb 17 at 10:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Théophile, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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
• 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). – Arthur Feb 17 at 1:50
• – Martin R Feb 17 at 8:31
• @Théophile: More Information here: How to search on this site? and here: Announcing a third-party search engine for Math StackExchange. – Martin R Feb 17 at 17:43

## 3 Answers

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