Question

If a² + b + c2=1 then the range of ab+bc+ca
is
1)[1, 0) 2)[71,-] 3)( - 4.4) 4)[7]​

Answer 1

Given,

$\longrightarrow a^2+b^2+c^2=1$

We need to find the range of $ab+bc+ca.$

Since $x^2\geq0$ for every real number $x,$ we see that,

$\longrightarrow (a+b+c)^2\geq0$

$\longrightarrow a^2+b^2+c^2+2(ab+bc+ca)\geq0$

$\longrightarrow 1+2(ab+bc+ca)\geq0$

$\longrightarrow ab+bc+ca\geq-\dfrac{1}{2}\quad\quad\dots(1)$

Also it is true that,

$\longrightarrow(a-b)^2+(b-c)^2+(c-a)^2\geq0$

[the sum of two or more non - negative real numbers is always non - negative.]

$\longrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\geq0$

$\longrightarrow a^2+b^2+c^2-ab-bc-ca\geq0$

$\longrightarrow1-ab-bc-ca\geq0$

$\longrightarrow ab+bc+ca\leq1\quad\quad\dots(2)$

Combining (1) and (2),

$\longrightarrow-\dfrac{1}{2}\leq ab+bc+ca\leq1$

Hence,

$\longrightarrow\underline{\underline{ab+bc+ca\in\left[-\dfrac{1}{2},\ 1\right]}}$