<s>[INST] Unique solutions for $\frac{A^2}{a}+\frac{B^2}{b}+\frac{C^2}{c}=1,\quad\text{and}\quad a+b+c=1.$ 
Let $A,B,C$ be strictly positive real numbers satisfying $A+B+C=1$ and let $a, b, c $ be real variables. Suppose $a, b, c$ satisfy the following system of equations:
$$\frac{A^2}{a}+\frac{B^2}{b}+\frac{C^2}{c}=1,\quad\text{and}\quad a+b+c=1.$$

It is clear that  $a=A$, $b=B$ and $c=C$ is one solution to the system. Can I conclude that the solution is unique and there exists no other solutions?
 [/INST] The solution is unique only if $a,b,c > 0$. This is because by Titu's lemma: $1 = \dfrac{A^2}{a}+\dfrac{B^2}{b}+\dfrac{C^2}{c} \ge \dfrac{(A+B+C)^2}{a+b+c}= \dfrac{1^2}{1} = 1$, and you have equality which means the $=$ must hold, and this occurs when $A = ak, B = bk, C = ck \implies k =1 \implies a = A, b = B, c = C$ as claimed. For if they are not required to be positive then there is another solution, namely: $a = \dfrac{A^2-B^2}{1-C^2}, b = \dfrac{B^2-A^2}{1-C^2},c = 1$.
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