Cubic Equation and Inequality

Well it is very obvious that $x = \lambda = \epsilon = \rho = -a$ . Also, we can deduce that $\lambda, \epsilon, \rho \in \mathbb{R}$ . So, we can apply the Cauchy Schwarz Inequality Theorem and Triangle Inequality Theorem of Vectors.

Thus, by Cauchy Schwarz Inequality

$|\vec{A} \cdot \vec{B}| \leq ||\vec{A}||||\vec{B}||$ ---> Q(1)

By Triangle Inequality Theorem

$||\vec{A} + \vec{B}|| \leq ||\vec{A}|| + ||\vec{B}||$ ---> Q(2)

Thus, $Q(1) + Q(2)$

$|\vec{A} \cdot \vec{B}| +||\vec{A} + \vec{B}|| \leq ||\vec{A}||||\vec{B}|| + ||\vec{A}|| + ||\vec{B}||$

Add both side by $1$ .

$|\vec{A} \cdot \vec{B}| +||\vec{A} + \vec{B}|| +1 \leq ||\vec{A}||||\vec{B}|| + ||\vec{A}|| + ||\vec{B}||+1$

Manipulate this!! Thus,...

$\frac{|\vec{A} \cdot \vec{B}| +||\vec{A} + \vec{B}|| +1}{||\vec{A}|| + 1} \leq ||\vec{B}|| +1$ . ---> Q(3)

Consider $\vec{A} = (0, \lambda, 0, \rho)$ and $\vec{B} =(3,3,3,3)$ . Thus, Q(3) will become....

$\frac{\sqrt{(\lambda +3)^2 + (\rho +3)^2 +18} + 3\lambda +3\rho +1}{\sqrt{\lambda^2 + \rho^2} +1} \leq 7$

At the same time, according to problem...

$\frac{\sqrt{(\lambda +3)^2 + (\rho +3)^2 +18} + 3\lambda +3\rho +1}{\sqrt{\lambda^2 + \rho^2} +1} \geq 7$

Therefore, we can deduce that

$\frac{\sqrt{(\lambda +3)^2 + (\rho +3)^2 +18} + 3\lambda +3\rho +1}{\sqrt{\lambda^2 + \rho^2} +1} = 7$

Substituting $\lambda = \rho = -a$ , the expression simplifies as $E = \frac{\sqrt{(6-a)^2+a^2}+6a+1}{\sqrt2 a + 1}.$ The square root is only defined for $0 \leq a \leq 6$ . It has its maximum value of 6 when $a = 0, 6$ and its minimum value of $3\sqrt 2$ when $a = 3$ . Let's first use the maximum value as upper bound: $E \leq E' = \frac{6+6a+1}{\sqrt2 a + 1} = \frac{6a+7}{\sqrt2 a + 1}.$ If $E \geq 7$ , then certainly $E' \geq 7$ . It is easy to check that neither numerator nor denominator has a zero on the interval $[0,6]$ , so that this is a monotonous function on that interval; it reaches its maximum and minimum value at the end points. $\text{min}(E') = E'_{a = 6} = 43/(6\sqrt 2 + 1) < 7;$ $\text{max}(E') = E'_{a = 0} = 7/1 = 7.$ Thus $E' \geq 7$ only if $a = 6$ ; and since we know that at $a = 6$ , $E = E'$ , we now also know that $E \geq 7$ only if $a = 6$ ; and in that case equality holds: the answer is $E = \boxed{7}$ .