MHB Show that 36≤4(s³+t³+u³+v³)-(s⁴+t⁴+u⁴+v⁴)≤48

[anemone]

Let the real numbers $s,\,t,\,u,\,v$ satisfy the relations $s+t+u+v=6$ and $s^2+t^2+u^2+v^2=12$.

Show that $36 \le 4(s^3+t^3+u^3+v^3)-(s^4+t^4+u^4+v^4) \le 48$.

 

[anemone]

Solution of other:

Spoiler

Observe that

$4(s^3+t^3+u^3+v^3)-(s^4+t^4+u^4+v^4)$

$=-((s-1)^4+(t-1)^4+(u-1)^4+(v-1)^4)+6(s^2+t^2+u^2+v^2)-4(s+t+u+v)+4$

$=-((s-1)^4+(t-1)^4+(u-1)^4+(v-1)^4)+52$

Now, introducing $x=s-1$, $y=t-1$, $z=u-1$, $t=v-1$, we need to prove the inequalities

$4\le x^4+y^4+z^4+t^4\le 16$,

under the constraint $x^2+y^2+z^2+t^2=(s^2+t^2+u^2+v^2)-2(s+t+u+v)+4=4$---(1)

Now, the rightmost inequality in (1) follows from the power mean inequality:

$x^4+y^4+z^4+t^4\ge \dfrac{(x^2+y^2+z^2+t^2)^2}{4}=4$

For the other one, expanding the brackets we note that

$(x^2+y^2+z^2+t^2)^2=(x^4+y^4+z^4+t^4)+q$ where $q$ is a non negative number, so

$x^4+y^4+z^4+t^4\le (x^2+y^2+z^2+t^2)^2=16$ and we're done.