$ a+b+c+d=6 , a^2+b^2+c^2+d^2=12$ $\implies$ $ 36 \leq 4(a^3+b^3+c^3+d^3)-(a^4+b^4+c^4+d^4) \leq48 $

Question

Let $a , b, c, d$ be real numbers such that $$ a+b+c+d=6 \\ a^2+b^2+c^2+d^2=12$$ How do we prove that $$ 36 \space \leq\space 4(a^3+b^3+c^3+d^3)-(a^4+b^4+c^4+d^4) \space\leq48\space$$ ?

Answer 1

lemma1 : $$0\le a,b,c,d\le 3$$

proof: Use Cauchy-Schwarz inequality,we have $$b^2+c^2+d^2\ge\dfrac{1}{3}(b+c+d)^2\Longrightarrow 12-a^2\ge \dfrac{(6-a)^2}{3}$$ so $$\Longrightarrow 0\le a\le 3$$

lemma 2: $$4a^3-a^4\ge 2a^2+4a-3,0\le a\le 3$$

proof: $$\Longleftrightarrow a^4-4a^3+2a^2+4a-3\le 0$$ since $$a^4-4a^3+2a^2+4a-3=(a-1)^2(a+1)(a-3)\le 0$$

**lemm3:**$$4a^3-a^4\le 4a^2$$

proof: since $$4a^3-a^4=4a^2-(a^2-2a)^2\le 4a^2$$

USe lemma 2: we have $$4(a^3+b^3+c^3+d^3)-(a^4+b^4+c^4+d^4)=\sum_{cyc}(4a^3-a^4)\ge\sum_{cyc}(2a^2+4a-3)=36$$

Use lemma3: we have $$4(a^3+b^3+c^3+d^3)-(a^4+b^4+c^4+d^4)=\sum_{cyc}(4a^3-a^4)\le \sum_{cyc}4a^2=48$$ By Done