if you suppose that all the $a_i$, all the $b_i$, and all the $c_i$ are distinct, can't you do that by induction ? One can assume wlog that $a_1 > \ldots > a_d > 0$ and $b_1 > \ldots > b_d > 0$ etc.. so that taking $p \to \infty$ you can see that $a_1+b_1=c_1$. Hence $$a_1(1+\sum_2^d $a_1\left(1+\sum_2^d \frac{a_k}{a_1}^p)^{\frac1p} left(\frac{a_k}{a_1}\right)^p\right)^{1/p} + b_1(1+\sum_2^d b_1\left(1+\sum_2^d \frac{b_k}{b_1}^p)^{\frac1p} left(\frac{b_k}{b_1}\right)^p\right)^{1/p} = c_1(1+\sum_2^d c_1\left(1+\sum_2^d \frac{c_k}{c_1}^p)^{\frac1p}$$ left(\frac{c_k}{c_1}\right)^p\right)^{1/p}$$ and a Taylor expansion for $p \to \infty$ tells you that $\frac{a_2}{a_1}=\frac{b_2}{b_1}=\frac{c_2}{c_1}$. Continuing this way, one can see that the only family of solutions is $A=\lambda C$ and $B=(1-\lambda) C$. Too simple to be true ?
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if you suppose that all the $a_i$, all the $b_i$, and all the $c_i$ are distinct, can't you do that by induction ? One can assume wlog that $a_1 > \ldots > a_d > 0$ and $b_1 > \ldots > b_d > 0$ etc.. so that taking $p \to \infty$ you can see that $a_1+b_1=c_1$. Hence $$a_1(1+\sum_2^d \frac{a_k}{a_1}^p)^{\frac1p} + b_1(1+\sum_2^d \frac{b_k}{b_1}^p)^{\frac1p} = c_1(1+\sum_2^d \frac{c_k}{c_1}^p)^{\frac1p}$$ and a Taylor expansion for $p \to \infty$ tells you that $\frac{a_2}{a_1}=\frac{b_2}{b_1}=\frac{c_2}{c_1}$. Continuing this way, one can see that the only family of solutions is $A=\lambda C$ and $B=(1-\lambda C)$B=(1-\lambda) C$. Too simple to be true ? |
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if you suppose that all the $a_i$, all the b_i, $b_i$, and all the $c_i$ are distinct, can't you do that by induction ? One can assume wlog that $a_1 > \ldots > a_d > 0$ and $b_1 > \ldots > b_d > 0$ etc.. so that taking $p \to \infty$ you can see that $a_1+b_1=c_1$. Hence $$a_1(1+\sum_2^d \frac{a_k}{a_1}^p)^{\frac1p} + b_1(1+\sum_2^d \frac{b_k}{b_1}^p)^{\frac1p} = c_1(1+\sum_2^d \frac{c_k}{c_1}^p)^{\frac1p}$$ and a Taylor expansion for $p \to \infty$ tells you that $\frac{a_2}{a_1}=\frac{b_2}{b_1}=\frac{c_2}{c_1}$. Continuing this way, one can see that the only familly family of solution solutions is $A=\lambda C$ and $B=(1-\lambda C)$. Too simple to be true ? |
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