Art of Problem Solving Forum

Altheman

Prove that in any triangle $ABC$ the following
inequality holds
$\cos A+\cos B+\cos C\ge\frac{1}{4}(3+\cos(A-B)+\cos(B-C)+\cos(C-A)).$

**Posted:** Tue Jul 21, 2009 8:10 pm

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Altheman's Problem Column

Dr Sonnhard Graubner

hello, we know that
$\cos(\alpha-\beta)=\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)=$
$\frac{(a^2+c^2-b^2)}{2ac}\frac{(a^2+b^2-c^2)}{2ab}+\frac{2A}{ac}\frac{2A}{ab}$
analogouosly we get
$\cos(\beta-\gamma)=\frac{(a^2+c^2-b^2)}{2ac}\frac{(a^2+b^2-c^2)}{2ab}+\frac{2A}{ac}\frac{2A}{ab}$
$\cos(\gamma-\alpha)=\frac{(a^2+b^2-c^2)}{2ab}\frac{(b^2+c^2-a^2)}{2bc}+\frac{2A}{ab}\frac{2A}{bc}$
inserting this in our inequality we get
$\displaystyle{\frac{3a^3b^2c+3ab^3c^2+3a^3bc^2-4abc^4-6a^2b^2c^2+3a^2bc^3- 4ab^4c+bc^5+}{2a^2b^2c^2}}$ +
$\displaystyle{\frac{+b^5c-2a^3c^3+a^5c+ac^5-2a^3b^3+ab^5+a^5b-4a^4bc+ 3ab^2c^3+3a^2b^3c-2b^3c^3}{2a^2b^2c^2}\geq0}$ .
Setting $a=y+z,b=x+z,c=x+y$ in the nominator we get after some algebra
$z^4(x-y)^2+y^4(x-z)^2+x^4(y-z)^2+2(x^3z^3+z^3y^3+y^3x^3-3x^2y^2z^2)\geq0$
this is true because of $z^4(x-y)^2+y^4(x-z)^2+x^4(y-z)^2\ge0$ and $x^3z^3+z^3y^3+y^3x^3\geq 3x^2y^2z^2$ is true because of AM-GM inequality.
A nice problem!
Sonnhard.

**Posted:** Sat Aug 15, 2009 6:06 am

me@home

real solution
real idea

**Posted:** Fri Oct 30, 2009 11:43 pm

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