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If in a triangle $ABC$ we have $$1=2\cos A\cos B\cos C+\cos A\cos B+\cos B\cos C+\cos C\cos A\ ,$$ then the triangle will be equilateral triangle.

I tried but except few steps,could not prove it. Putting $A=B=C=\frac{\pi}{3}$ makes both sides equal. How should i prove it?Please help me.

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  • $\begingroup$ DO you remember the source of such a nice problem? $\endgroup$ – lab bhattacharjee Jan 7 '16 at 16:45
  • $\begingroup$ Thanks & here is another solution. $\endgroup$ – lab bhattacharjee Jan 7 '16 at 17:50
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Observe that the equality fails if one of the $\cos$ is negative. We can assume that triangle is acute. By Jensen and followed by AM-GM inequality: $x = \cos A, y = \cos B, z = \cos C\to x+y+z \leq \dfrac{3}{2}$: $1 = 2xyz + xy+yz+zx \leq 2\cdot \dfrac{(x+y+z)^3}{27}+\dfrac{(x+y+z)^2}{3}\leq \dfrac{2}{27}\cdot \left(\dfrac{3}{2}\right)^3+\dfrac{1}{3}\cdot \left(\dfrac{3}{2}\right)^2=1\to $ equality must occur when $x = y = z \to $ triangle is equilateral.

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For brevity write $a=\cos A$, $b=\cos B$, $c=\cos C$. Then we have $$C=\pi-A-B$$ and so $$c=-\cos(A+B)=-ab+\sin A\sin B$$ and so $$(c+ab)^2=(1-a^2)(1-b^2)\ .$$ Expanding, rearranging and using the identity you are given, $$a^2+b^2+c^2=1-2abc=ab+bc+ca\ .$$ Now $$(a-b)^2+(b-c)^2+(c-a)^2=2a^2+2b^2+2c^2-2ab-2bc-2ca$$ and so $$(a-b)^2+(b-c)^2+(c-a)^2=0\ ,$$ which shows that $a,b,c$ are all equal; hence $A,B,C$ are all equal.

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As $A+B+C=\pi,\cos(A+B)=\cdots=-\cos C$

and using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,

$$2\cos A\cos B\cos C=\cos C[\cos(A-B)+\cos(A+B)]$$

$$=-\cos(A+B)\cos(A-B)-\cos^2C$$

$$=-(\cos^2A-\sin^2B)-\cos^2C$$

$$=1-(\cos^2A+\cos^2B+\cos^2C)$$

So, the equation reduces to $$\sum(\cos A-\cos B)^2=0$$

Can you take it from here?

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