$a;b;c\geq 0$ such that : $a^2+b^2+c^2=1$.
Prove : $0\leq ab+bc+ca-2abc \leq \frac{7}{27}$
Thanks :)
P/s : I have no ideas about this problem :(
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$\begingroup$ This is as saying that their harmonic mean lies in between $\dfrac{10}9$ and $\dfrac32$ . Also, try expanding $(a+b+c)^2$ $\endgroup$– LucianMay 22, 2014 at 9:04
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$\begingroup$ are you sure the right side is OK? $c=0,ab \le 0.5$, but $0.5> \frac{7}{27}$ $\endgroup$– chenbaiMay 22, 2014 at 12:59
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2 Answers
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Negative!
Try $a=b=c=\dfrac{1}{\sqrt{3}}$
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$\begingroup$ @Awesome : Of course.... :D $\endgroup$– user87543May 22, 2014 at 10:15
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$\begingroup$ @Awesome Have you even tried substituting that into the expression? The inequality and everything else seems to hold in your case. The equality is reached as well (i.e. $ab+bc+ca-2abc=0$). $\endgroup$ May 22, 2014 at 13:22
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This is the incorrect problem. The correct problem is
Given $a,b,c \ge 0, a+b+c = 1,$ prove $0 \le ab+bc+ca-2abc \le \frac{7}{27}$.
For the LHS, $ab \ge abc$ and so on so we are done.
For the RHS, $\sqrt{xy} \le x+y = 1-z $ then $z(x+y)+xy(1-2z) \le z(1-z)+\frac{(1-2z)(1-z)^2}4 \le \frac{7}{27}$ by calculus methods.
answered May 22, 2014 at 12:47
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1$\begingroup$ +1 for correcting the question and doing it... Like a boss... $\endgroup$ May 22, 2014 at 14:00