Cool Inequality 6 - Making it Really Tough

since Nobody has posted a solution I will post it now for Pi Han Goh .

We will prove the inequality

$\Large{36\le 4\left( { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }+{ d }^{ 3 } \right) -\left( { a }^{ 4 }+{ b }^{ 4 }+{ c }^{ 4 }+{ d }^{ 4 } \right) \le 48}$

Observe that

$\Large{4\left( { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }+{ d }^{ 3 } \right) -\left( { a }^{ 4 }+{ b }^{ 4 }+{ c }^{ 4 }+{ d }^{ 4 } \right) =-\left( \left( a-1 \right) ^{ 4 }+{ \left( b-1 \right) }^{ 4 }+{ \left( c-1 \right) }^{ 4 }+{ \left( d-1 \right) }^{ 4 } \right) +6\left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 } \right) -4\left( a+b+c+d \right) +4\\ =-\left( \left( a-1 \right) ^{ 4 }+{ \left( b-1 \right) }^{ 4 }+{ \left( c-1 \right) }^{ 4 }+{ \left( d-1 \right) }^{ 4 } \right) +52}$

Now, introducing $x=a-1, y=b-1, z=c-1, t=d-1$ we need to prove

$\Large{16\ge { x }^{ 4 }+{ y }^{ 4 }+{ z }^{ 4 }+{ t }^{ 4 }\ge 4}$

under the constrain

$\Large{{ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }+{ t }^{ 2 }=\left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 } \right) -2\left( a+b+c+d \right) +4=4}$

Now clearly

$\Large{{ x }^{ 4 }+{ y }^{ 4 }+{ z }^{ 4 }+{ t }^{ 4 }\ge \frac { \left( { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }+{ t }^{ 2 } \right) ^{ 2 } }{ 4 } =4}$

For other one on expanding we note that

$\Large{{ \left( { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }+{ t }^{ 2 } \right) ^{ 2 } }=\left( { x }^{ 4 }+{ y }^{ 4 }+{ z }^{ 4 }+{ t }^{ 4 } \right) +q}$

Where $q$ is a non-negative number, so

$\Large{\left( { x }^{ 4 }+{ y }^{ 4 }+{ z }^{ 4 }+{ t }^{ 4 } \right) \le { \left( { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }+{ t }^{ 2 } \right) ^{ 2 } }=16}$

and we are done!