# Inequality

• January 1st 2008, 10:22 AM
james_bond
Inequality
Given that $a+b\le c+1$, $b+c \le a+1$, $c+a\le b+1$ for the numbers $a,b,c \ge0$, show that $a^2+b^2+c^2\le2abc+1$.
• January 2nd 2008, 06:00 AM
colby2152
I am interested in this question. Squaring out some terms, I get the following inequality:

$2(ab+ac+bc)-3 \le a^2 + b^2 + c^2$

Definitely not what we are looking for!!:confused:
• January 2nd 2008, 07:07 AM
Isomorphism
Quote:

Originally Posted by colby2152
I am interested in this question. Squaring out some terms, I get the following inequality:

$2(ab+ac+bc)-3 \le a^2 + b^2 + c^2$

Definitely not what we are looking for!!:confused:

Do you want a solution?

We deduce the problem is equivalent to
$(a+b+c+1)^2 \leq 2(1+a)(1+c)(1+b)$

This is a much better form :)
• January 2nd 2008, 12:43 PM
JaneBennet
Notice that the inequality we need to prove contains a term in $abc$. Thus we have to try somehow or other to bring that term into our working.

Now, since we are working with non-negative numbers, we can multiply all the inequalities together. Let’s try that.

$(a+b)(b+c)(c+a)\ \leq\ (a+1)(b+1)(c+1)$

The LHS is equal to $(ab+bc+ca)(a+b+c)-abc$ (expand them out and check). The RHS is equal to $abc+(ab+bc+ca)+(a+b+c)+1$. Hence

$(ab+bc+ca)(a+b+c)\ \leq\ 2abc+(ab+bc+ca)+(a+b+c)+1$

Which doesn’t bring us any closer to a solution, but at least we now have the $abc$ term to work with. :rolleyes:
• January 4th 2008, 11:16 AM
Opalg
I'm sure this is supposed to be a slick application of one of those Jensen-type inequalities that they drill math olympiad candidates with. Here's a much more pedestrian approach (aesthetically unappealing because it breaks the symmetry).

If $a+b\leqslant c+1$ and $c+a\leqslant b+1$ then $b-c\leqslant 1-a$ and $c-b\leqslant 1-a$. Thus $|b-c|\leqslant 1-a$. In particular, $1-a\geqslant0$. Also, $(b-c)^2\leqslant (1-a)^2$. Then

. . . . . . $\begin{array}{rcl}a^2+b^2+c^2 -2abc &=&
a^2 + \frac12(1+a)(b-c)^2 + \frac12(1-a)(b+c)^2 \vspace{1ex} \\
&\leqslant&a^2 + \frac12((1+a)(1-a)^2 + \frac12(1-a)(1+a)^2 \vspace{1ex} \\ &=&a^2 + \frac12((1-a^2)(1-a) + \frac12(1-a^2)(1+a) \vspace{1ex} \\
&=& a^2 + 1 - a^2 = 1.\end{array}$
• January 5th 2008, 07:03 PM
ThePerfectHacker
Quote:

Originally Posted by Opalg
I'm sure this is supposed to be a slick application of one of those Jensen-type inequalities that they drill math olympiad candidates with.

Using Jensen's inequality is recent.
• January 5th 2008, 08:31 PM
Isomorphism
Quote:

Originally Posted by Opalg
I'm sure this is supposed to be a slick application of one of those Jensen-type inequalities that they drill math olympiad candidates with. Here's a much more pedestrian approach (aesthetically unappealing because it breaks the symmetry).

If $a+b\leqslant c+1$ and $c+a\leqslant b+1$ then $b-c\leqslant 1-a$ and $c-b\leqslant 1-a$. Thus $|b-c|\leqslant 1-a$. In particular, $1-a\geqslant0$. Also, $(b-c)^2\leqslant (1-a)^2$. Then

. . . . . . $\begin{array}{rcl}a^2+b^2+c^2 -2abc &=&
a^2 + \frac12(1+a)(b-c)^2 + \frac12(1-a)(b+c)^2 \vspace{1ex} \\
&\leqslant&a^2 + \frac12((1+a)(1-a)^2 + \frac12(1-a)(1+a)^2 \vspace{1ex} \\ &=&a^2 + \frac12((1-a^2)(1-a) + \frac12(1-a^2)(1+a) \vspace{1ex} \\
&=& a^2 + 1 - a^2 = 1.\end{array}$

That is a brilliant solution, Opalg