Question
If $a, b, c$ are distinct positive real numbers and $a^{2}+b^{2}$ $+c^{2}=1$, then $a b+b c+c a$ is(A) less than 1(B) equal to 1(C) greater than 1(D) any real number
Step 1
Therefore, we can write the inequality $a^{2}+b^{2}-2ab \geq 0$ or $2ab \leq a^{2}+b^{2}$. This inequality can be derived using the arithmetic mean is greater than or equal to the geometric mean as well. Show more…
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