we have $36\le 4\sum a^3 -\sum a^4\le 48$ iff $16\ge \sum (a-1)^4\ge 4$. now substitute $x=a-1,y=b-1,z=c-1$ and $t=d-1$. we have $x+y+z+t=2$ and $x^2+y^2+z^2+t^2=4$ and we must prove $16\ge \sum x^4 \ge 4$. by CS we have $4(\sum x^4)\ge (\sum x^2)^2=16$, so $\sum x^4\ge 4$. on the other hand we have $16=(\sum x^2)^2=\sum x^4 + 2\sum x^2y^2 \ge \sum x^4$, hence the result. by $[3,1,1,1]$ we get the lower bound, and by $[2,2,2,0]$ we get the upper bound.

It's also India 2011.

I have posted the solution here: http://www.math.vn/showthread.php?t=14929 (didn't know that it is an IMO shortlist problem).

There is an alternative version of this problem. Prove that $a^4+b^4+c^4+d^4-12abcd \leq 48$ (with same conditions).

orl wrote: Let the real numbers $a,b,c,d$ satisfy the relations $a+b+c+d=6$ and $a^2+b^2+c^2+d^2=12.$ Prove that \[36 \leq 4 \left(a^3+b^3+c^3+d^3\right) - \left(a^4+b^4+c^4+d^4 \right) \leq 48.\] Proposed by Ukraine Generalization of RHS is also true: For $a,b,c,d \in \mathbb{R}$ that satisfy $a^2+b^2+c^2+d^2=12$, prove: \[4 \left(a^3+b^3+c^3+d^3\right) - \left(a^4+b^4+c^4+d^4 \right) \leq 48 \]

Your conjecture does not work, try $(3, \sqrt{3}, 0, 0)$.

can_hang2007 wrote: I have posted the solution here: http://www.math.vn/showthread.php?t=14929 (didn't know that it is an IMO shortlist problem). That doesn't work.

One half of this inequality is quite easy: Obviously, for any real number $x$, $4x-x^{2}<=4$ (as their difference is $-(x-2)^{2}$). So, $4(a^{3}+b^{3}+c^{3}+d^{3})-a^{4}-b^{4}-c^{4}-d^{4} = a^2(4a-a^{2}) + b^{2} ( 4b-b^{2}) + c^2(4c-c^2) +d^2(4d-d^2) \leq 4a^2+4b^2+4c^2+4d^2 = 48$, hence one part is shown. For the second part, $4(a^3+b^3+c^3+d^3) = 4(a+b+c+d)(a^2+b^2+c^2+d^2) - 4(ab(a+b)+...) = 4*6*12-4(ab(a+b)+...)$ and $a^4+b^4+c^4+d^4 = (a^2+b^2+c^2+d^2)^2-2(a^2b^2+...) = 12^2-2(a^2b^2+...)$. So, our goal is to prove the following: $2(ab(a+b)+...)+54 \geq (a^2b^2+...)$, or equivalently, $54+(ab(ab-2a-2b)+...) \geq 0$, or $54+(ab(ab-2a-2b+4)+...) \geq 4(ab+...) $, but $ab+... = ((a+b+c+d)^2-(a^2+b^2+c^2+d^2))/2 = 24$. So, we need to prove that $ (ab(a-2)(b-2)+...) \geq -6$, or $((a-1)^2-1)((b-1)^2-1)+...) \geq -6$, or $(((a-1)(b-1))^2+...)+12-2((a-1)^2+...) \geq -6$, but it's the same as $(((a-1)(b-1))^2+...)>=0$, which is obviously true. It shows that the equality is achieved if and only if three of the numbers $a,b,c,d$ is equal to $1$, and the fourth one is $3$.

kunny wrote: can_hang2007 wrote: I have posted the solution here: http://www.math.vn/showthread.php?t=14929 (didn't know that it is an IMO shortlist problem). That doesn't work. maybe only vietname can visit it ...resently

Nak wrote: Your conjecture does not work, try $(3, \sqrt{3}, 0, 0)$. Please check your computations before you post something. $4\cdot 3^3- 3^4 + 4\cdot3^{\frac{3}{2}}-3^2=18+12\sqrt{3} \approx 38.7846$ and I think it's perfectly true.

Ok, so I quote it here: can_hang2007 wrote: Lời giải: Sử dụng bất đẳng thức Cauchy-Schwarz, ta có $b^2+c^2+d^2 \ge \frac{(b+c+d)^2}{3}.$ Từ đây suy ra $12-a^2 \ge \frac{6-a)^2}{3}.$ Giải bất phương trình này, ta được $0 \le a \le 3.$ Chứng minh tương tự, ta cũng có 0 \le a,\;b,\;c,\;d \le 3. Từ kết quả trên, ta có $(a-1)^2(a+1)(a-3) \le 0.$ Mà $(a-1)^2(a+1)(a-3)=a^4-4a^3+2a^2+4a-3,$ nên $4a^3-a^4 \ge 2a^2+4a-3.$ Cộng bất đẳng thức này với ba đánh giá tương tự, ta được ${4(a^3+b^3+c^3+d^3) -(a^4+b^4+c^4+d^4) \ge 2(a^2+b^2+c^2+d^2)+4(a+b+c+d)-12=2\cdot 12+4\cdot 6-12 =36.}$ Lại có $4a^3-a^4 =4a^2-(a^2-2a)^2 \le 4a^2,$ nên $4(a^3+b^3+c^3+d^3)-(a^4+b^4+c^4+d^4) \le 4(a^2+b^2+c^2+d^2)=4\cdot 12=48.$ Bài toán chứng minh xong.

Cám ơn, can_hang2007.　

Official Solutions.

Solution 2 is the same as Can's. Congratulations!

MathUniverse wrote: Please check your computations before you post something. I am sorry, your conjecture is true. The same idea as in original problem works. WLOG $3 \geq a \geq b \geq c \geq d \geq 0$; $a^2 +b^2+c^2+d^2=12$; $a+b+c+d=x$; $4 \left(a^3+b^3+c^3+d^3\right) - \left(a^4+b^4+c^4+d^4 \right) = S$; $a_1=a-1, b_1=b-1, c_1=c-1, d_1=d-1$. Then $a_1+b_1+c_1+d_1=x-4$; $a_1^2+b_1^2+c_1^2=d_1^2=16-2x \Longrightarrow x \leq 8$; $a_1^4+b_1^4+c_1^4+d_1^4=-S-4x+76 \Longrightarrow S= 76-4x-(a_1^4+b_1^4+c_1^4+d_1^4)$; $a_1^4+b_1^4+c_1^4+d_1^4=(16-2x)^2-2(a_1^2b_1^2+ \cdots + c_1^2d_1^2) \leq (16-2x)^2-2a_1^2(b_1^2+c_1^2+d_1^2)$ $S \geq 36 \Longleftrightarrow 40 - 4x \geq a_1^4+b_1^4+c_1^4+d_1^4 \Longleftarrow$ $\Longleftarrow 40 -4x \geq (16-2x)^2-2(a_1^2b_1^2+ \cdots + c_1^2d_1^2) \Longleftarrow a_1^2(b_1^2+c_1^2+d_1^2) \geq 2(6-x)(9-x)$. Now it is enough to prove that $a_1^2(b_1^2+c_1^2+d_1^2) \geq 2(6-x)(9-x)$. Consider following cases: 1. $x \geq 6$ 2. $x \in (5, 6)$ and $b_1^2+c_1^2+d_1^2 \geq 4$ 3. $x \in (5, 6)$ and $b_1^2+c_1^2+d_1^2 \leq 4$ 4. $x \leq 5$ First case: $x \leq 8 \Longrightarrow 2(6-x)(9-x) \leq 0$. Second case: suppose that $c_1^2 >1$ then $c_1 > 1$ because $c_1 = c-1 \geq -1 \Longrightarrow a \geq b \geq c > 2 \Longrightarrow a^2 +b^2 +c^2+d^2 > 12$ - contradiction $\Longrightarrow c_1^2, d_1^2 \leq 1 \Longrightarrow b_1^2 \geq 2, a_1^2 \geq 2 \Longrightarrow a_1^2(b_1^2+c_1^2+d_1^2) \geq 8$. Now it is enough to prove that $4 \geq (6-x)(9-x)$ which is true because $6-x \leq 1, 9-x \leq 4$. Third case: $4 \geq a_1^2, b_1^2+c_1^2+d_1^2 \Longrightarrow a_1^2(b_1^2+c_1^2+d_1^2) \geq 4(16-2x-4) =8(6-x)$. Now it is enough to prove that $4(6-x) \geq (6-x)(9-x)$ which is true because $9-x \leq 4$. Fourth case: $a_1, b_1, c_1, d_1 \in [-1, 2], a_1 + b_1 +c_1+d_1 = x-4 \leq 1 \Longrightarrow$ $b_1, c_1, d_1 \in [-1, 1] \Longrightarrow a_1^4+b_1^4+c_1^4+d_1^4 \leq 3 + 16 \leq 20 \leq 40 - 4x$. The conclusion follows. [moderator edit: topic cleared.]

@Nak how did you get this part: Nak wrote: $ 4\geq a_{1}^{2}, b_{1}^{2}+c_{1}^{2}+d_{1}^{2}\Longrightarrow a_{1}^{2}(b_{1}^{2}+c_{1}^{2}+d_{1}^{2})\geq 4(16-2x-4) =8(6-x) $

Because $ (4, 16-2x-4) \succeq (a_{1}^{2}, b_{1}^{2}+c_{1}^{2}+d_{1}^{2})$.

Very nice solution @Nak, much better than mine.

Lagrange multiplier method let $L(a,b,c,d,s,t)=4\sum a^3-\sum a^4+s(\sum a-6)+t(\sum a^2-12)$ by derivatives $12a^2-4a^3+s+2ta=0$(b,c,d hold analogously) by subtracting $(a-b)[6(a+b)-2(a^2+ab+b^2)+t]=0$ so either a=b,or b=c,or $3(a+b)-(a^2+ab+b^2)=-\frac{u}{2}=3(a+c)-(a^2+ac+c^2)$ hence b=c or a+b+c=3 if a=b=c=3 then d=3 hence a=b=c=1 obtaining minimum 36 otherwice,there exists two equal numbers assuming b=c(we can also suppose none of them equals 3) by discussing with a,b,d as above,we get a=d or b=a or b=d,yielding (a,b,c,d)=(x,x,y,y) or (x,y,y,y) by substituting we get another extremum 48.

orl wrote: Let the real numbers $a,b,c,d$ satisfy the relations $a+b+c+d=6$ and $a^2+b^2+c^2+d^2=12.$ Prove that \[36 \leq 4 \left(a^3+b^3+c^3+d^3\right) - \left(a^4+b^4+c^4+d^4 \right) \leq 48.\] Proposed by Nazar Serdyuk, Ukraine $$3(a^2+b^2+c^2) \ge (a+b+c)^2$$$$3(12-d^2) \ge (6-d)^2$$$$3d \ge d^2$$$$a,b,c,d \in [0,3]$$$$ \sum (4a^3 -a^4) \leq 48= 4 (\sum a^2)$$$$0 \leq \sum (a^2(a-2)^2)$$$$a=x+1,b=y+1,c=z+1,d=t+1$$$$(a-1)^4= x^4 =a^4-4a^3+6a^2-4a+1$$$$\sum x^4=-\sum (4a^3 -a^4)+6 \cdot 12 - 4 \cdot 6+4 =-\sum (4a^3 -a^4)+52$$$$\sum (4a^3 -a^4) = 52- \sum x^4$$$$36 \leq 52- \sum x^4$$$$\sum x^4 \leq 16=(x^2+y^2+z^2+t^2)^2$$equality case $(1,1,1,3)$

Bruh. Note that $-a^4+4a^3=-(a-1)^4+6a^2-4a+1$ thus \[4\left(a^3+b^3+c^3+d^3\right) - \left(a^4+b^4+c^4+d^4 \right)=-(a-1)^4-(b-1)^4-(c-1)^4-(d-1)^4+6(a^2+b^2+c^2+d^2)-4(a+b+c+d)+4.\]This implies we need to prove: $4\le (a-1)^4+(b-1)^4+(c-1)^4+(d-1)^4\le 16.$ Substitute $w=a-1,x=b-1,y=c-1,z=d-1.$ We want to prove $4\le w^4+x^4+y^4+z^4\le 16$ given $w+x+y+z=2$ and $w^2+x^2+y^2+z^2=a^2+b^2+c^2+d^2-2a-2b-2c-2d+4=4.$ The right bound can easily be noticed as we square $w^2+x^2+y^2+z^2$ to get \[w^4+x^4+y^4+z^4=16-2w^2x^2-2w^2y^2-2w^2z^2-2x^2y^2-2x^2z^2-2y^2z^2\le 16.\]The left bound can also easily be realized because $w^4+x^4+y^4+z^4\ge \frac{(w^2+x^2+y^2+z^2)^2}{4}\ge 4$ as desired.

Set $(a-1,b-1,c-1,d-1)=(a',b',c',d')$ then it suffices to show: $$16\geq \sum_{\text{cyc}} a'^4\geq 4$$The left inequality is: $$16= \sum_{\text{cyc}} (a'^2)^2\geq \sum_{\text{cyc}} a'^4$$The right is true by Power Mean.

Add $-52=-6 \sum a^2 + 4 \sum a - 4$ to both sides of the inequality and negate it to get that the inequality is equivalent to $$4 \le \sum ((a-1)^2)^2 \le 16.$$Let $(w,x,y,z)=((a-1)^2,(b-1)^2,(c-1)^2,(d-1)^2)$ so that $w,x,y,z \ge 0$ and $w+x+y+z=4$. We have that $$4=\frac{(w+x+y+z)^2}{4} \le \sum w^2 \le \sum w^2 + 2 \sum wx = (w+x+y+z)^2 =16,$$where the left follows from Cauchy and the right follows from the fact that $w,x,y,z \ge 0$ so we are done.

Let $w = a-1, x = b-1, y = c-1, z = d-1$. The inequality is equivalent to $4 \le \sum w^4 \le 16$, while the two relations are equivalent to $\sum w = 2$ and $\sum w^2 = 4$. By Cauchy-Schwarz we have $$(w^4+x^4+y^4+z^4)(1+1+1+1)\ge (w^2+x^2+y^2+z^2)^2 \implies \sum w^4 \ge 4$$with equality when $(w, x, y, z) = (1, 1, 1, -1)$ and also $$16 = (w^2+x^2+y^2+z^2)^2 = \sum w^4+2(w^2x^2+w^2y^2+w^2z^2+x^2y^2+x^2z^2+y^2z^2) \ge \sum w^4$$with equality when $(w, x, y, z) = (2, 0, 0, 0)$. For the original inequality, the upper bound occurs when $(a, b, c, d) = (2, 2, 2, 0)$ and the lower bound when $(a, b, c, d) = (3, 1, 1, 1)$.

1) $4\sum a^3-\sum a^4 \le 48$ By $C-S\implies$ $4\sum a^3 \le 4\sqrt{(\sum a^4)(\sum a^2)}=4\sqrt {12\sum a^4}=2\sqrt{48\sum a^4}$ Since $a^4,b^4,c^4,d^4 \ge 0$ We can use AM-GM So $2\sqrt{48\sum a^4} \le \sum {a^4} + 48$ $\blacksquare$ 2) $4\sum a^3-\sum a^4 \ge 36$ We have $\sum ({a^2-2a}) =0 \implies \sum {(a-1)^2}=4$ Also $\sum(4a^3-a^4)=\sum(6a^2-4a+1-(a-1)^4)=52-\sum(a-1)^4$ Hence we need to show $\sum(a-1)^4 \le 16\implies \sum(a-1)^4 \le (\sum(a-1)^2)^2$ and it is always true $\blacksquare$ 3)The upper bound occurs when $(a, b, c, d) = (2, 2, 2, 0)$ and the lower bound when $(a, b, c, d) = (3, 1, 1, 1)$ so we are done

BCW equality :heart_eyes: Let $S=\sum_{\text{cyc}} 4a^3-a^4$. Claim: $S \leq 48$. Proof. We have $$0 \leq \sum_{\text{cyc}} a^2(a-2)^2 = \sum_{\text{cyc}} (a^4-4a^3+4a^2) = 48-S \rightarrow S \leq 48. \square $$ Now, note that $$3(b^2+c^2+d^2) \geq (b+c+d)^2 \rightarrow 0 \leq a \leq 3,$$which implies $a,b,c,d \in [0,3]$. Claim: $S \geq 36$. Proof. Notice that we have $$0 \geq \sum_{\text{cyc}} (a+1)(a-1)^2(a-3) = \sum_{\text{cyc}} (a^4-4a^3+2a^2+4a-3)=36-S \rightarrow S\geq 36.\square$$ Equality occurs at $(2,2,0,0)$ for the upper bound and $(3,3,1,1)$ for the lower bound (and permutations). $\blacksquare$

The equality cases are at $(a, b, c, d) = (2, 2, 2, 0)$ for the upper bound and $(3, 1, 1, 1)$ for the lower bound. Actually, this is all the "information" we really need: Upper Bound: Sum cyclically the inequality $x^2(x-2)^2 \geq 0$. This yields $$\sum_{\mathrm{cyc}} (4a^3-a^4) \leq \sum_{\mathrm{cyc}} 4a^2 = 48.$$ Lower Bound: First we establish a bound on the variables. Using C-S, $$(b+c+d)^2 \leq 3(b^2+c^2+d^2) \iff (6-a)^2 \leq 3(12-a^2) \iff a \in [0, 3],$$where the inequality is sharp at both ends. This allows us to sum cyclically the estimate $(x-1)^2(x-3)(x+1)\leq 0$ to get $$\sum_{\mathrm{cyc}}(4a^3-a^4) \geq \sum_{\mathrm{cyc}} 2a^2+4a - 3 = 36.$$

Through the binomial theorem and substituting the given values in, the inequality we are asked to prove is equivalent to showing \[ 4 \leq (a-1)^4 + (b-1)^4 + (c-1)^4 + (d-1)^4 \leq 16.\] We have the equations \[ (a-1)+(b-1)+(c-1)+(d-1) = 2\]and \[ (a-1)^2 + (b-1)^2 + (c-1)^2 + (d-1)^2 = 4.\] Firstly, to prove the lowerbound of $\sum (a-1)^4$, we use Cauchy Schwarz: \begin{align*} \left( (a-1)^4 +(b-1)^4 + (c-1)^4 + (d-1)^4 \right) \left( \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} \right) & \geq \left( \frac{1}{2} \left( \sum (a-1)^2 \right) \right) \\ & = \left( \frac{1}{2} \cdot 4 \right)^2 \\ &= 4. \end{align*} Now we prove the upperbound of $16$. Note that $w^2+x^2+y^2+z^2 \leq (w+x+y+z)^2$ for nonnegative $w$, $x$, $y$ and $z$. Substituting $w = (a-1)^2$, $x = (b-1)^2$, $y = (c-1)^2$ and $z = (d-1)^2$ gives \begin{align*} (a-1)^4 + (b-1)^4 + (c-1)^4 + (d-1)^4 & \leq \left( (a-1)^2 + (b-1)^2 + (c-1)^2 + (d-1)^2 \right)^2 \\ & = 16. \end{align*}

From walkthrough Equality cases occur at $(2, 2, 2, 0)$ and $(3, 1, 1, 1)$. Now consider, \begin{align*} (x-2)^2x^2 &\geq 0 \end{align*}This expands as, \begin{align*} (x^2 - 4x + 4)x^2 &\geq 0\\ \iff x^4 - 4x^3 + 4x^2 &\geq 0 \end{align*}Now consider summing cyclically to find, \begin{align*} 4(a^2 + b^2 + c^2 + d^2) &\geq 4(a^3 + b^3 + c^3 + d^3) - (a^4 + b^4 + c^4 + d^4)\\ \iff 48 &\geq 4(a^3 + b^3 + c^3 + d^3) - (a^4 + b^4 + c^4 + d^4) \end{align*}Thus proving the upper bound, alongside the equality cases. Now to prove the lower bound note that, \begin{align*} (b+c+d)^2 &\leq 3(b^2+c^2+d^2)\\ (6-a)^2 &\leq 3(12 - a^2)\\ 36 - 12a + a^2 &\leq 36 - 3a^2\\ 0 &\leq 12a - 4a^2\\ 0 &\leq a(3 - a) \end{align*}Hence $a \in [0, 3]$, with similar results for $b$, $c$ and $d$. Then consider, \begin{align*} (x+1)(x-1)^2(x-3) &\leq 0 \end{align*}This expands as, \begin{align*} (x^2 - 2x + 1)(x^2 - 2x - 3) &\leq 0\\ x^4 - 4x^3 + 2x^2 + 4x - 3 &\leq 0 \end{align*}Summing cyclically we find, \begin{align*} 2(a^2 + b^2 + c^2 + d^2) + 4(a+b+c+d) - 12 &\leq 4(a^3 + b^3 + c^3 + d^3) - (a^4 + b^4 + c^4 + d^4)\\ 36 &\leq 4(a^3 + b^3 + c^3 + d^3) - (a^4 + b^4 + c^4 + d^4) \end{align*}as desired.

that is so tricky Let $S$ be the quantity. First, some bounding. Notice $b+c+d=6-a$ and $b^2+c^2+d^2=12-a^2$. Note both quantities are positive. Then \[36-3a^2=3(b^2+c^2+d^2)\ge (|b|+|c|+|d|)^2\ge (b+c+d)^2=a^2-12a+36\implies 12a\ge 4a^2\]thus $a\in [0,3]$ and similarly for the other variables. Next note $4a^2\ge 4a^3-a^4$ thus \[48=4(a^2+b^2+c^2+d^2)\ge S.\] Now notice \[2a^2+4a-3\le 4a^3-a^4\iff (a-3)(a-1)^2(a+1)\le 0\implies 36=2\cdot 12+4\cdot 6-12\le S\]and we are done. $\blacksquare$

Here's a solution along with (detailed) intuition: Let: $$S=4(a^3+b^3+c^3+d^3)-(a^4+b^4+c^4+d^4).$$For upper-bound, notice that $(a, b, c, d) = (2, 2, 2, 0)$ gives the upper bound. Here we have terms of the form: $a^4-4a^3$. We have to create sum of squares involving $(a-2), a$ factors. Notice that: $$a^2(a-2)^2 = \sum a^4 - 4 \sum a^3 + 4\sum a^2 = -S +48 \ge 0$$This gives the upper bound $S \le 48$. For lower-bound, we proceed with a similar approach on the problem. Notice that: $a+b+c+d=6$ and $a^2+b^2+c^2+d^2=12.$ It hints for Cauchy-Schwarz (Essentially 1978 USAMO P1). Notice that: $$(b^2+c^2+d^2)(3) \ge (b+c+d)^3.$$$$\implies (12-a^2)(3) \ge (6-a)^2$$$$\implies a \le 3.$$ So, we got the information that $a \le 3$. Now, notice that: $(a, b, c, d) = (3, 1, 1, 1)$ gives the lower-bound. But, now we dont need a $(a-3)^2$ term, instead it suffice to have a linear $(a-3)$ term. Now, notice that: $$(a-3)(a-1)^2 = a^3-5a^2+7a-3.$$Now, we need to multiply by a linear factor to get $a^4-4a^3$ thing. Multiplying by $(a-\alpha)$ with $a^3-5a^2+7a-3$: $$(a - \alpha) (a^3-5a^2+7a-3) = a^4-4a^3+\cdots$$It gives away $\alpha = -1$. Thus: $$\sum (a-3)(a-1)^2(a+1) \le 0.$$$$\sum a^4-4a^3 \le 36.$$

Let $f(a,b,c,d)=4(a^3+b^3+c^3+d^3)-(a^4+b^4+c^4+d^4)$ and $$S=\left \{(a,b,c,d)\in \left [-\sqrt{12},\sqrt{12} \right]^4: a+b+c+d=6, a^2+b^2+c^2+d^2=12 \right \}.$$It is clear that $S$ is compact and so $f$ attains a global minimum and maximum over this set. We can use Lagrange Multipliers. $$\nabla f(a,b,c,d)=\lambda_1 \nabla (a+b+c+d)+\lambda_2 \nabla (a^2+b^2+c^2+d^2) \implies 4x^3-12x^2+2\lambda_2 a+\lambda_1=0, \qquad x=a,b,c,d \qquad (1)$$As $a,b,c,d$ satisfy the same cubic equation, two of them must be equal; wlog assume that $a=d$. If $a,b,c$ are the three roots of the cubic in $(1)$ then $a+b+c=\frac{12}{4}=3$ by Vieta's relations. We also have $2a+b+c=6$, $2a^2+b^2+c^2=12$. Solving this system we can find that there are no solutions. Thus either $a=c$ or $b=c$ (or $a=b$ but this is essentially the same as $a=c$). If $a=c$ then $3a+b=6$, $3a^2+c^2=12$. Solving we get $(a,b)\in \{(1,3),(2,0)\}$. If $b=c$ then $2a+2b=6$, $2a^2+2b^2=12$ and we find $(a,b)=\left (\frac{3\pm \sqrt 3}2,\frac{3\mp \sqrt 3}2\right)$. So overall the minimum and maximum is attained for $$(a,b,c,d)\in \left \{(1,3,1,1),(2,0,2,2),\left (\frac{3\pm \sqrt 3}2,\frac{3\mp \sqrt 3}2, \frac{3\mp \sqrt 3}2, \frac{3\pm \sqrt 3}2\right)\right \}.$$ We can calculate that $f(1,3,1,1)=36$, $f(2,0,2,2)=48$, $f\left (\frac{3\pm \sqrt 3}2,\frac{3\mp \sqrt 3}2, \frac{3\mp \sqrt 3}2, \frac{3\pm \sqrt 3}2\right)=45$. Hence the minimum of $f$ is $36$ and the maximum is $48$, as desired.

bruh The problem is equivalent to proving $-16 \le -(1 + 1 + 1 + 1) + 4(a + b + c + d) - 6(a^2 + b^2 + c^2 + d^2) + 4(a^3 + b^3 + c^3 + d^3) - (a^4 + b^4 + c^4 + d^4) \le -4$, or letting $w = a - 1$ and cyclic variants, its equivalent to $w + x + y + z = 2, w^2 + x^2 + y^2 + z^2 = 4$, and proving $4 \le w^4 + x^4 + y^4 + z^4 \le 16$. The lower bound follows from Cauchy Schwarz, we see $(w^4 + x^4 + y^4 + z^4)(1 + 1 + 1+ 1) \ge (w^2 + x^2 + y^2 + z^2)^2 = 16$, so we get $w^4 + x^4 + y^4 + z^4 \ge 4$. The upper bound follows from $w^4 + x^4 + y^4 + z^4 \le (w^2 + x^2 + y^2 + z^2)^2 = 16$.