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It is another problem from AoPS: Let $a,b,c,d\geq 0$ with $a^2+b^2+c^2+d^2=4$. Then $$\frac{28}{\sqrt3}\geq a^3+b^3+c^3+d^3+3\left(a+b+c+d\right)+2\sqrt3 (7-4\sqrt3)\left(abcd-\frac{1}3\right).$$

I rewrote it as $$24 a b c d + 14 \sqrt3\geq a^3 + b^3+c^3+d^3+14 \sqrt3 a b c d+3(a+b+c+d)$$ but now what can I do? I tried Lagrange but the stationary points have no closed form.

  • This question just made me chuckle. – marty cohen Jan 15 '20 at 23:48
  • @martycohen Maybe this answer of mine will also make you chuckle – Maximilian Janisch Jan 15 '20 at 23:59
  • $24 - 14\sqrt{3} = (8\sqrt 3 - 14)\sqrt3$, and $8\sqrt{3}$ and $14$ are the values of the function $f(a, b, c, d) = a^3 + b^3 + c^3 + d^3 + 3(a + b + c + d)$ at $(\sqrt{3}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$ and $(2, 0, 0, 0)$ respectively which do have relevance to the Lagrangian of $f(a, b, c, d) = a^3 + b^3 + c^3 + d^3 + 3(a + b + c + d)$ for the constraint $g(a, b, c, d) = a^2 + b^2 + c^2 + d^2 - 4$ –  Jan 16 '20 at 02:01

1 Answers1

1

The AoPS inequality: $$ \frac{28}{\sqrt3}\geq a^3+b^3+c^3+d^3+3\left(a+b+c+d\right)+2\sqrt3 (7-4\sqrt3)\left(abcd-\frac{1}3\right) \tag{$1$} $$ is equivalent to $$ 14\sqrt 3\color{red}{-8} \geq a^3+b^3+c^3+d^3+3\left(a+b+c+d\right)+(14\sqrt3-24)abcd\ . \tag{$1'$} $$ Which is false. Numerically, using sage:

sage: GG(a,b,c,d) = 28/sqrt(3) - (a^3+b^3+c^3+d^3) - 3*(a+b+c+d) - 2*sqrt(3)*(7-4*sqrt(3))*(a*b*c*d-1/3)                             
sage: A, B, C, D = 0.66, 0.66, 0.66, sqrt(4-3*0.66^2)                                                                                
sage: A, B, C, D                                                                                                                     
(0.660000000000000, 0.660000000000000, 0.660000000000000, 1.64109719395287)
sage: A^2+B^2+C^2+D^2                                                                                                                
4.00000000000000
sage: GG(A, B, C, D).n()                                                                                                             
-0.0142154410781897

Let us consider now the other inequality: $$ 14\sqrt 3 \geq a^3+b^3+c^3+d^3+3\left(a+b+c+d\right)+(14\sqrt3-24)abcd\ , \tag{$2$} $$ which is generously true, and insert some words about it.

We will get simple maximization inequalities for the separate expressions $abcd$, $a+b+c+d$, $a^3+b^3+c^3+ d^3$ under the constraint $a^2+b^2+c^2+d^2=4$. The first two expressions are controlled in a simple manner: $$ \begin{aligned} abcd &= \sqrt{(abcd)^2}= \sqrt{a^2b^2c^2d^2} \\ &\le \sqrt{\frac 14(a^2+b^2+c^2+d^2)}=1\ , \\[3mm] (a+b)^2 &\le 2(a^2+b^2)\\ (c+d)^2 &\le 2(c^2+d^2)\\ (a+b+c+d)^2 &\le 2(a+b)^2+2(c+d)^2\\ &\le 4(a^2+b^2)+4(c^2+d^2)\\ &=16\ ,\text{ so} \\ a+b+c+d&\le 4\ . \end{aligned} $$ The expression $a^3+b^3+c^3+d^3$ becomes maximal in one of the corners, since applying Lagrange multiplicators, using the function $$ F(a,b,c,d;t)=(a^3+b^3+c^3+d^3)-t(a^2+b^2+c^2+d^2-4) $$ leads to a system with solutions either at the boundary ($a=0$ or $b=0$ or $c=0$ or $d=0$), or satisfying $a=b=c=d$, but the point $a=b=c=d=1$ is the minimal value. We want the maximal value. Let us look then closer at the boundary $d=0$. This is the "same", with one less variable. Again we get critical interior points for $a=b=c$, but at this place we have a minimum, a.s.o. - We finally get the maximal value $8=2^3+0^3+0^3+0^3$ for $a,b,c,d$ taking the values $2,0,0,0$ in a suitable order.

So the given inequality is generously satisfied. To have a rough orientation, let us type: $$ \begin{aligned} &(a^3+b^3+c^3+d^3)+3(a+b+c+d)+ \underbrace{(14\sqrt 3-24)}_{\in (1/5,\ 1/4)}abcd \\ &\qquad\le 8+3\cdot 4 + (14\sqrt 3-24) \\ &\qquad=14\sqrt 3\color{blue}{-4} \\ &\qquad<14\sqrt 3\ . \end{aligned} $$

Note: I wanted to introduce some comments, but each of them was rather offensive.

dan_fulea
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