If a, b, c are distinct positive real numbers | vedclass

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(A) Given that $a, b, c$ are distinct positive real numbers such that $a^2+b^2+c^2=1$. We know the algebraic identity: $(a+b+c)^2 = a^2+b^2+c^2+2(ab+b

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less than $1$

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(A) Given that $a, b, c$ are distinct positive real numbers such that $a^2+b^2+c^2=1$. We know the algebraic identity: $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$. Since $a, b, c > 0$,we have $(a+b+c)^2 > a^2+b^2+c^2$,which implies $a^2+b^2+c^2+2(ab+bc+ca) > a^2+b^2+c^2$. This simplifies to $2(ab+bc+ca) > 0$,which is always true for positive numbers. Also,we know that $(a-b)^2 + (b-c)^2 + (c-a)^2 > 0$ because $a, b, c$ are distinct. Expanding this,we get $2(a^2+b^2+c^2) - 2(ab+bc+ca) > 0$. Substituting $a^2+b^2+c^2=1$,we get $2(1) - 2(ab+bc+ca) > 0$,which means $2 > 2(ab+bc+ca)$,or $ab+bc+ca Thus,the value of $ab+bc+ca$ is less than $1$.

If $a, b, c$ are distinct positive real numbers and $a^2+b^2+c^2=1$,then the value of $ab+bc+ca$ is

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