I've been having some trouble with this particular problem. Could someone provide a hint of some sort? My original attempt that turned out to be useless is to start off with $$(b+c)^2 + (a+c)^2 + (a+b)^2 = (b+c)^2 + (a+c)^2 + (a+b)^2$$ and subtract from both sides until I get the above inequality, but I found out I'd end up subtracting that very same inequality and would be right back where I started.
Prove for all real numbers $a, \ b,$ and $c$, $bc + ac + ab \leq a^2 + b^2 + c^2$.
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$\begingroup$
algebra-precalculus
proof-writing
3 Answers
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BIG HINT: $$2a^2+2b^2+2c^2\geq 2ab+2bc+2ac\\a^2+b^2+b^2+c^2+a^2+c^2\geq 2ab+2bc+2ac$$
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2HINT? No BIG HINT! haha – 2017-02-11
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0Question - as long as you show a statement implies something that is universally correct, then the statement itself is correct? – 2017-02-11
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0@AndrewTawfeek You would need to reverse the order; show that something which is universally correct implies your statement,or that those two statements are equivalent.Because something false can imply something true. – 2017-02-11
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0Thank you for your reply! I was wondering that regarding the other answer, but I appreciate you clearing up my confusion. :) – 2017-02-11
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$ a^2 + b^2 + c^2 \ge ab + bc + ca\\ \Longleftrightarrow \frac12(2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca) \ge 0\\ \Longleftrightarrow (a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 - 2ac + a^2) \ge 0 \\ \Longleftrightarrow (a-b)^2 +(b-c)^2 +(c-a)^2 \ge 0 $
which is obviously true. Giving a hint for this problem is as good as solving it entirely. :-)
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1Hey maybe you should've used $\iff$ instead of $\implies$ because something false can imply something true. – 2017-02-11
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0@kingW3, noted. Thanks. :-) – 2017-02-11
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внимание: overkill. Stating that for any $(a,b,c)\in\mathbb{R}^3$ we have $$ Q(a,b,c) = a^2+b^2+c^2-ab-ac-bc \geq 0 \tag{1}$$ is the same as stating that the following symmetric matrix $$ Q = \begin{pmatrix}1& -\frac{1}{2}&-\frac{1}{2}\\-\frac{1}{2}&1&-\frac{1}{2}\\-\frac{1}{2}&-\frac{1}{2}& 1\end{pmatrix}\tag{2}$$ is positive semi-definite. That is simple to prove through Sylvester's criterion (the determinants of the principal minors are $\geq 0$) or by computing the eigenvalues of $Q$, namely $\left\{0,\frac{3}{2},\frac{3}{2}\right\}$.