Min + Max? #1

• Find $m$

Notice that

${ \left( a+b+c+d \right) }^{ 2 }\ge 0\\ \Longrightarrow 1+2\left( ab+ac+ad+bc+bd+cd \right) \ge 0\\ \Longrightarrow ab+ac+ad+bc+bd+cd\ge -\frac { 1 }{ 2 } \\ \Longrightarrow K\ge -\frac { 1 }{ 2 } +2cd$ .

Also, ${ \left( c+d \right) }^{ 2 }+{ a }^{ 2 }+{ b }^{ 2 }\ge 0\Longrightarrow 2cd\ge -\left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 } \right) =-1$

Hence, $K\ge -\frac { 1 }{ 2 } +2cd\ge -\frac { 3 }{ 2 }$ .

$K=-\frac { 3 }{ 2 }$ when $a=b=0, c=-d, { c }^{ 2 }+{ d }^{ 2 }=1$ .

So ${ K }_{ min }=-\frac { 3 }{ 2 }$ when $(a,b,c,d)=(0,0,\pm \frac { 1 }{ \sqrt { 2 } } ,\mp \frac { 1 }{ \sqrt { 2 } } )$ .

• Find $M$

By applying the popular inequality $2ab\le { a }^{ 2 }+{ b }^{ 2 }$ , with some real positive $x$ we will have

$2xK=\left( 2ab \right) x+2a\left( xc \right) +2a\left( xd \right) +2b\left( xc \right) +2b\left( xd \right) +\left( 2cd \right) \left( 3x \right) \\ \le \left( { a }^{ 2 }+{ b }^{ 2 } \right) x+{ a }^{ 2 }+{ \left( xc \right) }^{ 2 }+{ a }^{ 2 }+{ \left( xd \right) }^{ 2 }+{ b }^{ 2 }+{ \left( xc \right) }^{ 2 }+{ b }^{ 2 }+{ \left( xd \right) }^{ 2 }+\left( { c }^{ 2 }+{ d }^{ 2 } \right) \left( 3x \right) \\ \Longrightarrow 2xK\le \left( x+2 \right) \left( { a }^{ 2 }+{ b }^{ 2 } \right) { + }{ \left( 2{ x }^{ 2 }+3x \right) }\left( { c }^{ 2 }+{ d }^{ 2 } \right)$

Now let choose $x>0$ such that $x+2=2{ x }^{ 2 }+3x\Longleftrightarrow { x }^{ 2 }+x-1=0\Longrightarrow x=\frac { \sqrt { 5 } -1 }{ 2 }$

Then

$2xK\le \left( x+2 \right) \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 } \right) =x+2\\ \Longrightarrow K\le \frac { x+2 }{ 2x } =\frac { \sqrt { 5 } +2 }{ 2 } .$

$K=\frac { \sqrt { 5 } +2 }{ 2 }$ when

$\begin{cases} a=b=xc=xd \\ x=\frac { \sqrt { 5 } -1 }{ 2 } \\ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }=1 \end{cases}\Longleftrightarrow \begin{cases} { a }^{ 2 }={ b }^{ 2 }=\frac { 5-\sqrt { 5 } }{ 20 } \\ { c }^{ 2 }={ d }^{ 2 }=\frac { 5+\sqrt { 5 } }{ 20 } \end{cases}$

So ${ K }_{ max }=\frac { 2+\sqrt { 5 } }{ 2 }$ when $\begin{cases} { a }={ b }=\pm \sqrt { \frac { 5-\sqrt { 5 } }{ 20 } } \\ { c }={ d }=\pm \sqrt { \frac { 5+\sqrt { 5 } }{ 20 } } \end{cases}$ .

Finally, $\left( m+M \right) = \frac { 2+\sqrt { 5 } -3 }{ 2 } =\boxed {\dfrac{ \sqrt { 5 } -1}2 } .$

Here is my alternative approach, using linear algebra:

$K = \frac12\left(\begin{array}{cccc} a & b & c & d \end{array}\right) \left(\begin{array}{cccc} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 3 \\ 1 & 1 & 3 & 0\end{array}\right) \left(\begin{array}{c} a \\ b \\ c \\ d \end{array}\right).$

We determine the eigenvalues of the matrix by solving $\left|\begin{array}{cccc} -\lambda & 1 & 1 & 1 \\ 1 & -\lambda & 1 & 1 \\ 1 & 1 & -\lambda & 3 \\ 1 & 1 & 3 & -\lambda\end{array}\right| = 0; \\ \lambda^4 - 14\lambda^2 - 16\lambda - 3 = 0; \\ (\lambda + 1)(\lambda^3 - \lambda^2 - 13\lambda - 3) = 0; \\ (\lambda + 1)(\lambda + 3) (\lambda^2 - 4\lambda - 1) = 0; \\ (\lambda + 1)(\lambda + 3) (\lambda - 2 + \sqrt{5}) (\lambda - 2 - \sqrt{5})= 0; \\ \lambda = -3,\ -1,\ 2-\sqrt{5},\ 2+\sqrt{5}.$ Thus we can rewrite $K$ using an eigenvalue basis as follows: $K = \tfrac12\left((-3)w^2 + (-1)x^2 + (2-\sqrt{5})y^2 + (2+\sqrt{5})z^2\right),$ where $(w,\dots z)$ are the $(a, \dots, d)$ expressed in a basis of orthonormal eigenvectors. We have $w^2 + x^2 + y^2 + z^2 = 1$ .

The least value of $K$ is obtained for a vector in the direction of the eigenvector belonging to the smallest eigenvalue, etc. Thus: $m = \tfrac12\cdot (-3)\cdot 1^2 = -1.5;\ \ \ M = \tfrac12\cdot (2+\sqrt{5}) \cdot 1^2 = 2.118; \\M + m = \frac{\sqrt{5}-1}2 = -1.5 + 2.118 = \boxed{0.618}.$