If $a an \alpha+\sqrt{a^2-1} an \beta+\sqrt{a^2+1} an \gamma=2 a$, where a is constant and $\a

Question

If $a 	an \alpha+\sqrt{a^2-1} 	an \beta+\sqrt{a^2+1} 	an \gamma=2 a$, where a is constant and $\alpha, \beta, \gamma$ are variable angles, then the least value of $3\left(	an ^2 \alpha+	an ^2 \beta+	an ^2 \gammaight)$ is $=$

Professor PC Thomas Classes · Answer

We have $$ \begin{aligned} & \left(a 	an \beta-\sqrt{a^2-1} 	an \alphaight)^2+\left(\sqrt{a^2+1} 	an \beta-\sqrt{a^2-1} 	an \gammaight)^2+\left(a 	an \gamma-\sqrt{a^2+1} 	an \alphaight)^2 \geq 0 \ & \Rightarrow\left\{a^2+a^2-1+a^2+1ight\}\left(	an ^2 \alpha+	an ^2 \beta+	an ^2 \betaight) \ & \Rightarrow 	an ^2 \alpha+	an ^2 \beta+	an ^2 \gamma \geq \frac{4}{3} \quad-\left\{a 	an \alpha+\sqrt{a^2-1} 	an \beta+\sqrt{a^2+1} 	an \gammaight\}^2 \geq 0 \end{aligned} $$