Double inequality

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Double inequality

Post Number:#1  Unread postby Katy729 » Fri Aug 04, 2017 1:21 pm

Prove that $0\le bc+ca+ab-2abc\le{7\over27}$, where $x,y$ and $z$ are non-negative real numbers satisfying $a+b+c=1$
Katy729
 
Posts: 34
Joined: Sat May 06, 2017 2:30 am

Re: Double inequality

Post Number:#2  Unread postby Katy729 » Sat Aug 19, 2017 11:10 pm

Someone? :(
Katy729
 
Posts: 34
Joined: Sat May 06, 2017 2:30 am

Re: Double inequality

Post Number:#3  Unread postby SMMamun » Tue Aug 22, 2017 10:04 pm

There is a mistake in the variables of the problem. I've corrected $ x, y, z$ respectively to $ a, b, c $.

We observe that
$ bc + ca + ab - 2abc $
$ = bc - abc + ca - abc + ab -abc + abc $
$ = bc(1-a) + ca(1-b) + ab(1-c) + abc $
$ = bc(b+c) + ca(c+a) + ab(a+b) + abc \geq 0 $, because $ a, b, c $ each are non-negative.
So left part of the inequality is proved. In fact, this part is obvious as soon as we compare and realize that $ ab \geq abc$ and so on, since $ a, b, c $ can at best be equal to $ 1 $.

Now we observe that
$(1-2a)(1-2b)(1-2c)$
$ = 1 - 2(a+b+c) + 4(bc+ca+ab) - 8abc $
$ = 1 - 2 + 4(bc+ca+ab) - 8abc$
$\therefore bc + ca + ab - 2abc=\frac {1}{4} (1-2a)(1-2b)(1-2c) + \frac{1} {4} $

Let's evaluate the sign and value of $(1-2a)(1-2b)(1-2c)$.
Case I: At best one of $a, b, c$ can be greater than $\frac{1}{2}$, and in that case, value of $(1-2a)(1-2b)(1-2c)$ will be negative.

Case II: $a, b, c \leq \frac{1} {2}$. In such case, by AM-GM inequality
$(1-2a)(1-2b)(1-2c)\leq \Big(\frac{(1-2a) + (1-2b) + (1-2c)} {3}\Big)^3$
$\Rightarrow(1-2a)(1-2b)(1-2c)\leq\frac{1}{27}$

In both cases,
$bc + ca + ab - 2abc \leq \frac{1} {4} \cdot \frac{1}{27} + \frac{1} {4}$
$\therefore bc + ca + ab - 2abc \leq \frac{7} {27} $
SMMamun
 
Posts: 49
Joined: Thu Jan 20, 2011 6:57 pm

Re: Double inequality

Post Number:#4  Unread postby Katy729 » Thu Aug 24, 2017 10:22 pm

Yes sMMamum, it was obviously an error.
Very good and clear solution.
Thanks!! :)
Katy729
 
Posts: 34
Joined: Sat May 06, 2017 2:30 am


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