Quadratic Equation and Inequality

The minimum value of x+ $\frac{1}{x}$ is 2 , and therefore maximum value of the given expression is 2 .

$λ=(-∞,-4]⋃[4,+∞) \Rightarrow \lambda^2 - 4 \geq 0$ . Notice that $\lambda^2 - 4$ is the discriminant of the equation $x^2+λx+1=0$ . So, we can deduce that $x$ is a real number.

Consider the $\vec{A} = (x, \lambda)$ and $\vec{B} = (\lambda, x)$ . Since $x$ and $\lambda$ is $\mathbb{R}$ so we can apply the Cauchy Schwarz Inequality on R² using standard inner product.

So, by Cauchy Schwarz Inequality...

$-1 \leq \frac{\vec{A} \cdot \vec{B}}{||\vec{A}||||\vec{B}||} \leq 1$

$\Rightarrow \frac{2\lambda x}{x^2 + \lambda^2} \leq 1$

$\Rightarrow \frac{4\lambda x}{x^2 + \lambda^2} \leq 2$

Therefore, $a = 2$ .