Question

If ${\alpha }^2+{\beta }^2+{\gamma }^2=1$, then highest integral value of $\alpha \beta +\beta \gamma +\alpha \gamma$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

Given, ${\alpha }^2+{\beta }^2+{\gamma }^2=1$ ………..$\left(1\right)$

Now, $(\alpha -\beta {)}^2+(\beta -\gamma {)}^2+(\alpha -\gamma {)}^2\ge 0$

$\Rightarrow {\alpha }^2+{\beta }^2-2\alpha \beta +{\beta }^2+{\gamma }^2-2\beta \gamma +{\alpha }^2+{\gamma }^2-2\alpha \gamma \ge 0$

$\Rightarrow 2{\alpha }^2+2{\beta }^2+2{\gamma }^2-2(\alpha \beta +\beta \gamma +\alpha \gamma )\ge 0$

$\Rightarrow 2({\alpha }^2+{\beta }^2+{\gamma }^2)-2(\alpha \beta +\beta \gamma +\alpha \gamma )\ge 0$

$\Rightarrow 2-2(\alpha \beta +\beta \gamma +\alpha \gamma )\ge 0$ [From $\left(1\right)$]

$\Rightarrow 2\{1-(\alpha \beta +\beta \gamma +\alpha \gamma \left)\right\}\ge 0$

$\Rightarrow 1-\alpha \beta +\beta \gamma +\alpha \gamma \ge 0$

$\therefore \alpha \beta +\beta \gamma +\alpha \gamma \le 1$

Hence the highest integral value of $\alpha \beta +\beta \gamma +\alpha \gamma$ is $1$.