Algebra is easy
Find the real value of k such that the system of equation a 2 + a b = k c 2 , b 2 + b c = k a 2 , c 2 + c a = k b 2 have positive real number solution for a , b and c .
The answer is 2.
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2 solutions
a = b = c ∧ a > 0 ∧ k = 2
You have demonstrated that k = 2 is a solution when a = b = c . But what if a = b = c is not fulfilled? Is there any other value of k that satisfy the given constraints?
No. The 3 equations are symmetrical. Therefore, a = b = c , rewriting one of the equations using a gives a 2 + a 2 = k a 2 ⇒ 2 a 2 = k a 2 ⇒ k = 2 . It is really that simple.
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No, by your logic, the only solutions to the system of equations ⎩ ⎪ ⎨ ⎪ ⎧ a = b c b = a c c = a b is a = b = c = 0 , 1 . But that is incorrect, because ( a , b , c ) = ( 1 , − 1 , − 1 ) , ( − 1 , 1 , − 1 ) , ( − 1 , − 1 , 1 ) are also solutions that doesn't satisfy the condition a = b = c .
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Go read the problem again. a,b and c are positive reals. By my logic, the only solution a=b=c=1 and that solves your problem.
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@A Former Brilliant Member – I don't think you're getting my point. Just because you are given a system of symmetric equations, that doesn't make all the variables equal to each other. I've demonstrated a counterexample to illustrate this.
Adding the given equations and rearranging, we get (k-1)( a 2 + b 2 + c 2 )=ab+bc+ca. Now, for positive reals a, b, c, a 2 + b 2 + c 2 >=ab+bc+ca. Therefore k-1=1 or k=2