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Let $a, b, c ∈ R$, $a^{2}+b^{2}+c^{2}=1$and $A = ab + bc + ca$. Then $A$:

(A) $-\tfrac{1}{2}< A < 1 $

(B).$ −1 < A < 1$

(C) $-\tfrac{1}{2}< A \leq 1 $

(D) $-\tfrac{1}{2}\leq A \leq 1 $

$$\left ( a+b+c \right )^{2}= \left ( \left ( a+b \right )+c \right )^{2}$$ $$\Rightarrow \left ( a^{2}+b^{2}+2ab \right )+c^{2}+ 2\left ( a+b \right )c$$ $$\Rightarrow \left ( a^{2}+b^{2}+c^{2} \right ) +2\left ( ab + bc + ca \right )$$ $$\Rightarrow 1+2A\geq 0$$

$$\Rightarrow A\geq -\frac{1}{2}$$

I got the left hand constraint of A but i don't know how to proceed and find the upper limit.

Is this correct? and so do i automatically select option (D)

Any help will be appreciated.

AskMSellen
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1 Answers1

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It is obvious that $A=1$ can be reached, simply set $a=b=c$. This already gets rid of option (A) and (B). Since (D) includes (C) (that is, if (C) is true then (D) must also be true), one of the answers must be (D). The question now is: can $A=\frac12$ occur?

Well, in the reasoning

\begin{align} (a+b+c)^2&\geq 0\\ a^2+b^2+c^2+2(ab+bc+ca)&\geq 0\\ 1+2A&\geq 0\\ A&\geq -\tfrac12\\ \end{align}

what do we need for equality? We merely need to have the first expression to have an equality, that is, $a+b+c=0$. Now we need to find $a,b,c$ with $a+b+c=0$ and $a^2+b^2+c^2=1$. Substitution yields $a^2+b^2+(-a-b)^2=1$ or

$$a^2+ab+(b^2-\tfrac12)=0$$

which is simply a quadratic in $a$ and thus solvable, making $A=\frac12$ possible. The only answer can be (D).

To give explicit $a,b,c$ with $A=-\frac12$: solving the quadratic yields

\begin{align} a&=\tfrac12-\tfrac 16\sqrt{3}\\ b&=\tfrac13\sqrt{3}\\ c&=-\tfrac12-\tfrac 16\sqrt{3} \end{align}

so that $a^2+b^2+c^2=1$ and $A=ab+bc+ca=-\frac12$.