If in $\Delta ABC$ $$\cos A+\cos B+\cos C=\frac{3}{2}$$prove that it is an equilateral triangle without using inequalities. I tried using the cosine rule as follows:
$$\frac{b^2+c^2-a^2}{2bc}+\frac{a^2+c^2-b^2}{2ac}+\frac{a^2+b^2-c^2}{2ab}=\frac{3}{2}$$ $\implies$
$$a(b^2+c^2-a^2)+b(a^2+c^2-b^2)+c(a^2+b^2-c^2)=3abc$$ $\implies$
$$(a+b+c)(a^2+b^2+c^2)=2(a^3+b^3+c^3)+3abc$$
Any clues as to how I can take it from here?
A trick I often use with these sorts of problems is this. Let $A = P+Q$ and $B=P-Q$, so that we want to show that $Q=0$.
$A+B+C = 180 \Rightarrow C = 180 - (A+B) = 180 - 2P$. The condition becomes $$\cos(P+Q)+\cos(P-Q)+\cos(180°-2P) = \frac{3}{2}$$
Using standard trig identities yields
$$2\cos(P)\cos(Q)-2\cos^2(P)+1 = \frac{3}{2}$$
$$\Rightarrow 4\cos^2(P) - 4\cos(Q)\cos(P) + 1 = 0$$
This is a quadratic in $\cos(P)$, and since it must have real solutions, its discriminant must be $\ge 0$. Hence,
$$16\cos^2(Q)-16 \ge 0$$
$ \Rightarrow \cos^2(Q) \ge 1$ $ \Rightarrow \cos^2(Q) = 1$ $ \Rightarrow \cos(Q) = \pm1$ $\Rightarrow Q = 0 $ or $180$. However, $A-B = 2Q$, and since $A$ and $B$ cannot differ by $360$, $Q$ must be $0$ $\Rightarrow A=B$.
Now, if we swap $B$ with $C$ and repeat the process we will get $A=C$.
Hence, $A=B=C$, and the triangle is equilateral.
A different kind of trick : I thought you will like it.
So multiply the above equation by $a,b,c$ and add these up. Next , use the identity that $c = a \cos B + b \cos A$, and likewise for the other sides. After pairing these, you get: () $$ \frac{3(a+b+c)}{2} = a \cos A + b \cos B + c \cos C + a + b +c $$ which then simplifies to: $$ \frac{a+b+c}{2} = a\cos A + b\cos B + c \cos C \leq \frac{(a+b+c)(\cos A + \cos B + \cos C)}3 $$
by Chebyshev's inequality. But then, equality is attained here : this happens only under certain conditions. I leave you to conclude.
