inequality [triangle, A >= B >= C ==> a (C-B) + ... >= 0]

ARTI · P versus NP Offline Joined: 15 Jan 2005 Posts: 47

A+B+C=180 and A >= B >= C
Prove that a(C-B) + b(A-C) + c(B-A) >= 0

**Posted:** Thu Feb 03, 2005 6:10 am

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darij grinberg

You didn't tell us what a, b, c are, but it's almost sure that a, b, c are the sidelengths of the triangle with the angles A, B, C, so that your question is:

Let A, B, C be the angles and a, b, c the corresponding sidelengths of a triangle. Assume that $A\geq B\geq C$ . Prove that $a\left(C-B\right)+b\left(A-C\right)+c\left(B-A\right)\geq 0$ .

Indeed, after the Extended Sine Law, we have a = 2R sin A, b = 2R sin B and c = 2R sin C, where R is the circumradius of our triangle. In other words, a = f(A), b = f(B) and c = f(C), where the function f(x) is defined by f(x) = 2R sin x. Now, since the function sin x is concave on the interval $\left[0;\;\pi\right]$ , and 2R is positive, the function f(x) = 2R sin x is also concave on the interval $\left[0;\;\pi\right]$ . This interval is, of course, the interval where the angles A, B, C lie. Thus, our problem will immediately follow from the following more general problem:

General problem. Let A, B, C be three arbitrary real numbers in an interval I, and let f(x) be a concave function on I. Assume that $A\geq B\geq C$ . Then, $f\left(A\right)\left(C-B\right)+f\left(B\right)\left(A-C\right)+f\left(C\right)\left(B-A\right)\geq 0$ .

The solution of this general problem is easy: Since $A\geq B\geq C$ , we have $B-C\geq 0$ , $A-C\geq 0$ and $A-B\geq 0$ , so that $\frac{B-C}{A-C}\geq 0$ and $\frac{A-B}{A-C}\geq 0$ . Also, clearly, $\frac{B-C}{A-C}+\frac{A-B}{A-C}=\frac{A-C}{A-C}=1$ . Thus, since the function f(x) is concave on the interval I, we have by the definition of a concave function

$\frac{B-C}{A-C}f\left(A\right)+\frac{A-B}{A-C}f\left(C\right)\leq f\left(\frac{B-C}{A-C}A+\frac{A-B}{A-C}C\right)$ ;

since $\frac{B-C}{A-C}A+\frac{A-B}{A-C}C=\frac{\left(B-C\right)A+\left(A-B\right)C}{A-C}=\frac{B\left(A-C\right)}{A-C}=B$ , this rewrites as

$\frac{B-C}{A-C}f\left(A\right)+\frac{A-B}{A-C}f\left(C\right)\leq f\left(B\right)$ .

Multiplication by A - C (remembering that $A-C\geq 0$ ) yields

$\left(B-C\right)f\left(A\right)+\left(A-B\right)f\left(C\right)\leq \left(A-C\right)f\left(B\right)$ .

Now, subtracting $\left(B-C\right)f\left(A\right)+\left(A-B\right)f\left(C\right)$ , we get

$0\leq \left(A-C\right)f\left(B\right)-\left(\left(B-C\right)f\left(A\right)+\left(A-B\right)f\left(C\right)\right)$
$= \left(A-C\right)f\left(B\right)+\left(C-B\right)f\left(A\right)+\left(B-A\right)f\left(C\right)$
$=f\left(A\right)\left(C-B\right)+f\left(B\right)\left(A-C\right)+f\left(C\right)\left(B-A\right)$ ,

and the general problem is solved. $\blacksquare$

darij

**Posted:** Thu Feb 03, 2005 10:36 am

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nickolas

The initial inequality is very easy $a(C-B)+b(A-C)+c(B-A)\geq c(C-B)+c(A-C)+c(B-A)=0$

EDIT: I am sorry, this is wrong.

**Posted:** Tue Feb 08, 2005 10:55 am