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hien Posts: 623 Location: ASEAN	Posted: May 05, 2007, 1:14 am • # 1 Let $A,B,C$ be three angles of triangle $ABC$ . Prove that: $(1-\cos A)(1-\cos B)(1-\cos C)\ge\cos A\cos B\cos C$

hipsishopsis

Posts: 43

Posted: May 05, 2007, 6:18 am • # 2

Hidden Text

First note that if one of the angles is $\ge \frac{\pi}{2}$ radians then the left side is $\ge 0$ and right side is $\le 0$ .

So let's assume they are all in the range $0 < A,B,C < \frac{\pi}{2}$ .

$(1-\cos A)(1-\cos B)(1-\cos C) \ge \cos A \cos B \cos C$

On the given range this is equivalent to:
$(\frac{1}{\cos A}-1)(\frac{1}{cos B}-1)(\frac{1}{\cos C}-1) \ge 1$ .

Now note that as $0 < \cos x < 1$ when $0 < x < \frac{\pi}{2}$ every term on the left side is $\ge 1$ .

scorpius119

Posts: 1678
Location: ..., PA
United States

Posted: May 05, 2007, 11:44 am • # 3

hipsishopsis wrote:

Hidden Text

That last part doesn't work; it only guarantees each term nonnegative.
Hidden Text

It is true that we can reduce this to acute triangles, because if right/obtuse,
$(1-\cos A)(1-\cos B)(1-\cos C)\geq 0\geq \cos A\cos B\cos C$
Otherwise, let $x=\cos A,y=\cos B,z=\cos C$ . Observe that one of $(2x-1)(2y-1),(2y-1)(2z-1),(2z-1)(2x-1)$ must be nonnegative (since they are three numbers with nonnegative product). WLOG let $(2x-1)(2y-1)\geq 0$ . Now the inequality is equivalent to
$(1-x)(1-y)\geq (xy+(1-x)(1-y))z$
From $\cos (A-B)\leq 1$ , we have $\cos (A-B)+\cos (A+B)\leq 1+\cos (A+B)$ , which becomes $z\leq 1-2xy$ . We can now multiply by the positive $xy+(1-x)(1-y)$ . It remains to prove
$(1-x)(1-y)\geq (xy+(1-x)(1-y))(1-2xy)$
Guess what? It turns out that their difference is $xy(2x-1)(2y-1)$ (that's what motivated the WLOG in the first place), so we're done.

hipsishopsis

Posts: 43

Posted: May 05, 2007, 12:03 pm • # 4

scorpius119 wrote:

hipsishopsis wrote:

Hidden Text

That last part doesn't work; it only guarantees each term nonnegative.

Oops, you're right - careless thinking here

arqady

Posts: 5983
Location: shevah school, tel-aviv
Israel

Posted: May 05, 2007, 12:46 pm • # 5

hien wrote:

Let $A,B,C$ be three angles of triangle $ABC$ . Prove that:
$(1-\cos A)(1-\cos B)(1-\cos C)\ge\cos A\cos B\cos C$

$(1-\cos A)(1-\cos B)(1-\cos C)\ge\cos A\cos B\cos C\Leftrightarrow$
$\Leftrightarrow(a+b-c)^{2}(a+c-b)^{2}(b+c-a)^{2}\geq(a^{2}+b^{2}-c^{2})(a^{2}+c^{2}-b^{2})(b^{2}+c^{2}-a^{2}).$
Let $\Delta ABC$ is acute angle triangle and use the following inequality:
$(a+b-c)^{2}(a+c-b)^{2}\geq(a^{2}+b^{2}-c^{2})(a^{2}+c^{2}-b^{2})$ and similar.

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Michael Rozenberg

Teki-Teki

Posts: 553
Location: Within 12600 km of you.

Posted: May 06, 2007, 4:30 am • # 6

hmm... if $xyz \ge 8$ does that imply $(x-1)(y-1)(z-1) \ge 1$ for $x,y,z > 1$ ?

if so then the inequality follows from $\cos{A}\cos{B}\cos{C}\le \frac{1}{8}$ which can be shown by Jensen's.

EDIT: never mind, the first assumption does not hold - consider $x = 7, y = z = 2\sqrt{\frac{2}{7}}$

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Mathematics, rightly viewed, possesses not only truth, but supreme beauty - a beauty cold and austere, like that of sculpture. - Bertrand Russell

April Posts: 1271 Location: Hanoi, Vietnam	Posted: May 06, 2007, 5:39 am • # 7 $\bullet$ If triangle $ABC$ is obtuse then the proposed inequality is true. $\bullet$ If triangle $ABC$ is acute, then we have: $\prod\cos A\le\prod (1-\cos A)\iff\prod\cos A(1+\cos A)\le\prod(1-\cos^{2}A)\\ \iff 8\prod\cos A\prod\cos^{2}\frac{A}{2}\le\p...$ which is trivial.

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