If a^2+b^2+c^2=1,where a, b, c in mathbb{R},t | vedclass

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(D) We know that $(a-b)^2+(b-c)^2+(c-a)^2 \geq 0$. Expanding this,we get $2(a^2+b^2+c^2) - 2(ab+bc+ca) \geq 0$. Since $a^2+b^2+c^2 = 1$,we have $2(1)

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$\{\frac{-1}{2}, 1\}$

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(D) We know that $(a-b)^2+(b-c)^2+(c-a)^2 \geq 0$. Expanding this,we get $2(a^2+b^2+c^2) - 2(ab+bc+ca) \geq 0$. Since $a^2+b^2+c^2 = 1$,we have $2(1) - 2(ab+bc+ca) \geq 0$,which implies $ab+bc+ca \leq 1$. Also,we know that $(a+b+c)^2 \geq 0$. Expanding this,we get $a^2+b^2+c^2 + 2(ab+bc+ca) \geq 0$. Substituting $a^2+b^2+c^2 = 1$,we get $1 + 2(ab+bc+ca) \geq 0$,which implies $ab+bc+ca \geq -\frac{1}{2}$. Therefore,the range of $ab+bc+ca$ is $[-\frac{1}{2}, 1]$. The set of extreme values is $\{\frac{-1}{2}, 1\}$.

If $a^2+b^2+c^2=1$,where $a, b, c \in \mathbb{R}$,then the set of extreme values of $ab+bc+ca$ is

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