# Prove (or disprove) that for all real numbers $a$, $b$, and $c$, $3a^2 + 3b^2 + 3c^2 \geq (a + b + c)^2$

The problem in my textbook is to prove this inequality, but I can't seem to make any sense of their answer, so either there's something I'm not seeing, or the authors have made a mistake. The answer they give is of the following nature: The quadratic mean-arithmetic mean inequality implies that the square root of one third of the quantity $$a$$ squared plus $$b$$ squared plus $$c$$ squared is greater than or equal to one third of $$a$$ plus $$b$$ plus $$c$$. Now the next step in the given solution is what confuses me. They then square both sides of the equation to yield one third of $$a$$ squared plus $$b$$ squared plus $$c$$ squared is greater than or equal to one ninth of the square of the number $$a$$ plus $$b$$ plus $$c$$. Why is this squaring warranted? $$x$$ greater than or equal to $$y$$ by itself does not imply that x squared is greater than or equal to $$y$$ squared. A counterexample: $$(x, y) = (1,-4)$$. $$|x|$$ greater than or equal to $$|y|$$ does imply $$x$$ squared is greater than or equal to y squared, however, but we don't have an inequality of that form. So my question is: Why can both sides be squared in the solution such that the inequality still holds?

• For some basic information about writing mathematics at this site see, e.g., here, here, here and here. Jul 24 at 12:34
• From How to ask a good question: "Your question should be clear without the title After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title." Jul 24 at 12:34
• As the quadratic mean inequality applies even when $a, b, c$ are positive, it is clearly strong enough to conclude for the case when any of them have different signs and hence the sum lesser. However it would have made sense to be explicit, like saying $\sqrt{\frac13(a^2+b^2+c^2)} \geqslant \frac13(|a|+|b|+|c|) \geqslant \frac13|a+b+c|$. Then the squaring is clearly permissible. Jul 24 at 13:06
• Apply the Cauchy-Schwarz inequality so that the constant $3$ shows up in a natural way. Jul 24 at 14:20

This is an alternative solution not the answer to the OP's particular question. $$a^2+b^2+c^2\ge ab+bc+ca\tag{1}$$ Try to prove this yourself using $$AM-GM$$ inequality or Cauchy Schwarz Inequality.
Now just multiply $$(1)$$ by $$2$$ on both sides of the inequality and then add $$a^2+b^2+c^2$$ to both sides of the inequality.