Prove : $0\leq ab+bc+ca-2abc \leq \frac{7}{27}$

Question

$a;b;c\geq 0$ 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 :(

This is as saying that their harmonic mean lies in between $\dfrac{10}9$ and $\dfrac32$ . Also, try expanding $(a+b+c)^2$ — Lucian, May 22 '14 at 09:04
are you sure the right side is OK? $c=0,ab \le 0.5$, but $0.5> \frac{7}{27}$ — chenbai, May 22 '14 at 12:59

score 6 · Answer 1 · answered May 22 '14 at 08:58

6

Negative!

Try $a=b=c=\dfrac{1}{\sqrt{3}}$

answered May 22 '14 at 08:58

2

$0,0,1$ is easier. – evil999man May 22 '14 at 09:33
@Awesome : Of course.... :D – May 22 '14 at 10:15
@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. $ab+bc+ca-2abc=0$). – user26486 May 22 '14 at 13:22
@mathh What have I done... – evil999man May 22 '14 at 13:32

score 4 · Answer 2 · answered May 22 '14 at 12:47

4

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 '14 at 12:47

Sandeep Silwal

1

+1 for correcting the question and doing it... Like a boss... – evil999man May 22 '14 at 14:00

2 Answers2