4

Let $\{a,b,c,d\}\subset[0,1]$. Prove that: $$a^4+b^4+c^4+d^4+a^2b^2+b^2c^2+c^2d^2+d^2a^2+8(1-a)(1-b)(1-c)(1-d)\geq1$$ I tried convexity, the substitution $a=\frac{x}{x+1}...$ and more, but without success.

Michael Rozenberg
  • 194,933

3 Answers3

1

Mathematica Resolve can prove (or verify) it although we do not see step-by-step solution. The code is given by

ClearAll[f, pos1];
f[a_, b_, c_, d_] := a^4 + b^4 + c^4 + d^4 + a^2*b^2 + b^2*c^2 + c^2*d^2 
                       + d^2*a^2 + 8*(1 - a)*(1 - b)*(1 - c)*(1 - d) - 1;
pos1 := a >= 0 && a <= 1 && b >= 0 && b <= 1 && c >= 0 && c <= 1 &&d >= 0 && d <= 1;
Resolve[ForAll[{a, b, c, d}, pos1, f[a, b, c, d] >= 0], Reals]

My machine runs the code in 645 seconds. So the step-by-step solution in Mathematica's innards is very, very long (perhaps hundreds or thousands of A4 pages).

Also, the inequality may be expressed as SOS (Sum of Squares), i.e., the form of $z^TQz$. However, the size of $Q$ is $> 200$. So I don't want to continue this way.

Hope to see elegant solutions.

River Li
  • 37,323
1

Derivative test shows that only 5 critical points in range $\{a,b,c,d\}\in[0,1]$, by solving the equation system:

$$\begin{cases}\dfrac{\partial{f}}{\partial{a}}=4a^3+2ab^2+2ad^2-8(1-b)(1-c)(1-d)=0\\\dfrac{\partial{f}}{\partial{b}}=4b^3+2bc^2+2ba^2-8(1-a)(1-c)(1-d)=0\\\dfrac{\partial{f}}{\partial{c}}=4c^3+2cd^2+2cb^2-8(1-a)(1-b)(1-d)=0\\\dfrac{\partial{f}}{\partial{d}}=4d^3+2da^2+2dc^2-8(1-a)(1-b)(1-c)=0\end{cases}$$

where $f=a^4+b^4+c^4+d^4+a^2b^2+b^2c^2+c^2d^2+d^2a^2+8(1-a)(1-b)(1-c)(1-d)-1$.

The Gröbner basis contains 9 polynomials, where most of them are too long to paste here. The last one is:

$$(d-1)(2d-1)^3(d^2+1)(d^2-d+1)(6d^6-36d^5+100d^4-92d^3+32d^2-16d+9)(243d^7-1404d^6+3519d^5-4633d^4+3353d^3-1308d^2+248d-16)(531441d^{12}-2834352d^{11}+6928416d^{10}-10786284d^9+12885804d^8-12958704d^7+10799190d^6-7210332d^5+4007584d^4-1997172d^3+856272d^2-260528d+39710)(59049d^{14}-196830d^{13}+329508d^{12}-753300d^{11}+1938519d^{10}-3324330d^9+3932675d^8-3583448d^7+2511045d^6-1299104d^5+537554d^4-191304d^3+44136d^2-3972d+132)=0$$

So we can solve d, then c and b and a. In range $\{a,b,c,d\}\in[0,1]$, the only 5 solutions are:

  • $f(1/2,1/2,1/2,1/2)=0$, the minimum
  • $f(0.925481...,0.123238...,0.475260...,0.123238...)=0.0584473...$ and 3 cyclics, maybe the local minima or saddle points.

I use SageMath and code returns immediately:

from sage.all import *

R = PolynomialRing(QQ, 'a, b, c, d', order = 'lex')
a, b, c, d = R.gens()
f = a4 + b4 + c4 + d4 + a2*b2 + b2*c2 + c2*d2 + d2*a2 + 8(1 - a)(1 - b)(1 - c)(1 - d) - 1
f_a, f_b, f_c, f_d = diff(f, a), diff(f, b), diff(f, c), diff(f, d)
B = R.ideal(f_a, f_b, f_c, f_d).groebner_basis()
for f in B:
    print(factor(f), '= 0')
print(len(B), 'equations')

UPDATE

I missed the minimum $f(0,0,0,1)$ and 3 cyclics that are not critical points but on boundary.

Boundaries we should check:

  • $d=0,1$
  • $d=0,1$ and $c=0,1$
  • $d=0,1$ and $b=0,1$ (cyclic but not symmetric)
  • $d=0,1$, $c=0,1$ and $b=0,1$
  • $d=0,1$, $c=0,1$, $b=0,1$ and $a=0,1$ (fs are constants)

So there are many critical points to solve.

For example, $d=0$ can get 3 polynomials, where the last one is:

$$(c^9+48c^7-110c^6+136c^5-60c^4-190c^3+320c^2-176c+32)(9c^{18}+204c^{16}+36c^{15}-2652c^{14}-1764c^{13}+7878c^{12}-7760c^{11}+155640c^{10}-422596c^9+365336c^8+108704c^7-1046316c^6+1989248c^5-1850664c^4+1040752c^3-528848c^2+242896c-51984)=0$$

There are 4 solutions and none of them are minimum.

$d=1$ can get 10 polynomials, where the last one is:

$$c(c^2+1)(2c^2+1)(3c^2+2)=0$$

The only one solution is $a=b=c=0$ and we can get the minimum $f(0,0,0,1)=0$.

auntyellow
  • 579
0

Here is a human verifiable proof.

The desired inequality is written as $$(a^4 + c^4) + (b^4 + d^4) + (a^2 + c^2)(b^2 + d^2) + 8(1 - a)(1 - c) \cdot (1 - b)(1 - d) \ge 1. \tag{1}$$

WLOG, assume that $a \le c$, $b \le d$, and $c \le d$.

We can prove that \begin{align*} &(a^4 + c^4) + (b^4 + d^4) + (a^2 + c^2)(b^2 + d^2) + 8(1 - a)(1 - c) \cdot (1 - b)(1 - d)\\ \ge{}& \frac{(a^2 + c^2)^2}{2} + (b^4 + d^4) + (a^2 + c^2)(b^2 + d^2) \\ &\qquad + 8\left(1 - \sqrt{\frac{a^2 + c^2}{2}}\right)^2 (1 - b)(1 - d). \tag{2} \end{align*} (The proof of (2) is given at the end.)

Thus, it suffices to prove that \begin{align*} &\frac{(a^2 + c^2)^2}{2} + (b^4 + d^4) + (a^2 + c^2)(b^2 + d^2)\\ &\qquad + 8\left(1 - \sqrt{\frac{a^2 + c^2}{2}}\right)^2 (1 - b)(1 - d) \ge 1. \tag{3} \end{align*}

Let $x = \sqrt{\frac{a^2 + c^2}{2}}$. Then $0 \le x \le 1$. (3) is written as $$2x^4 + (b^4 + d^4) + 2x^2(b^2 + d^2) + 8(1 - x)^2 (1 - b)(1 - d) \ge 1. \tag{4}$$

Let $p := 2 - b - d, q := (1 - b)(1 - d)$ and $u := 1- x$. Then $p^2 \ge 4q$. (4) is written as \begin{align*} &{p}^{4}+2\,{p}^{2}{u}^{2}+2\,{u}^{4}-4\,{p}^{3}-4\,{p}^{2}q-4\,{p}^{2} u-4\,p{u}^{2}+4\,q{u}^{2}-8\,{u}^{3}+8\,{p}^{2}\\ &\qquad +12\,pq+8\,pu+2\,{q}^{2 }+8\,qu+16\,{u}^{2}-8\,p-16\,q-16\,u+7\\ &\ge 0. \tag{5} \end{align*}

From $p^2 \ge 4q$, let $q := \frac{p^2}{4 + s}$ for $s \ge 0$, (5) is equivalently written as $$As^2 + Bs + C \ge 0 \tag{6}$$ where \begin{align*} A &:= {p}^{4}+2\,{p}^{2}{u}^{2}+2\,{u}^{4}-4\,{p}^{3}-4\,{p}^{2}u-4\,p{u}^{2 }-8\,{u}^{3}\\ &\qquad +8\,{p}^{2}+8\,pu+16\,{u}^{2}-8\,p-16\,u+7, \\ B &:= 4\,{p}^{4}+20\,{p}^{2}{u}^{2}+16\,{u}^{4}-20\,{p}^{3}-24\,{p}^{2}u-32 \,p{u}^{2}\\ &\qquad -64\,{u}^{3}+48\,{p}^{2}+64\,pu+128\,{u}^{2}-64\,p-128\,u+56, \\ C &:= 2\,{p}^{4}+48\,{p}^{2}{u}^{2}+32\,{u}^{4}-16\,{p}^{3}-32\,{p}^{2}u-64 \,p{u}^{2}-128\,{u}^{3}\\ &\qquad +64\,{p}^{2}+128\,pu+256\,{u}^{2}-128\,p-256\,u +112. \end{align*}

We can prove that $A, B, C \ge 0$. Indeed, we have $$A = 2(1 - u)^2 + \frac12(1 - p)^4 + \frac12(p^2 + 2u^2 - 2p - 4u + 3)^2 \ge 0,$$ and \begin{align*} B &= 16\, \left( u-1 \right) ^{4}+ \left( 20\,{p}^{2}-32\,p+32 \right) \left( u-1 \right) ^{2}+16\,{p}^{2} \left( u-1 \right)\\ &\qquad + 4\,{p}^{4}-20 \,{p}^{3}+44\,{p}^{2}-32\,p+8\\ &\ge \left( 20\,{p}^{2}-32\,p+32 \right) \left( u-1 \right) ^{2}+16\,{p}^{2} \left( u-1 \right)\\ &\qquad + 4\,{p}^{4}-20 \,{p}^{3}+44\,{p}^{2}-32\,p+8\\ &\ge 0, \end{align*} (Note: We use $4 \cdot \left( 20\,{p}^{2}-32\,p+32 \right)\cdot (4\,{p}^{4}-20 \,{p}^{3}+44\,{p}^{2}-32\,p+8) \ge (16p^2)^2$ (easy).) and $$C = 8(1 - 2pu)^2 + 8(p + 2u - 2)^2 + 2(p^2 + 4u^2 - 4p - 8u + 6)^2 \ge 0.$$

We are done.

Proof of (2).

We have \begin{align*} &\mathrm{LHS}_{(2)} - \mathrm{RHS}_{(2)}\\ ={}& \frac{(a^2 - c^2)^2}{2} + 8\left((1-a)(1-c) - \left(1 - \sqrt{\frac{a^2 + c^2}{2}}\right)^2\right)(1 - b)(1 - d). \end{align*} We only need to prove the case that $$(1-a)(1-c) - \left(1 - \sqrt{\frac{a^2 + c^2}{2}}\right)^2 < 0.$$ Using $(1 - b)(1 - d) \le (1 - 0)(1 - (a+c)/2)$, it suffices to prove that $$\frac{(a^2 - c^2)^2}{2} + 8\left((1-a)(1-c) - \left(1 - \sqrt{\frac{a^2 + c^2}{2}}\right)^2\right)(1 - (a+c)/2) \ge 0. \tag{A1}$$

Let $p = a + c, q = ac$. We have $p \le 2$ and $p^2 \ge 4q$.

(A1) is written as $$\frac{p^2(p^2 - 4q)}{2} + 8\left(1 - p + q - \left(1 - \sqrt{\frac{p^2 - 2q}{2}}\right)^2\right)(1 - p/2) \ge 0$$ or $$f(q) := 8(2-p)\sqrt{\frac{p^2 - 2q}{2}} + (-2p^2 - 8p + 16)q + \frac12p^4 - 8p + 2p^3 \ge 0. \tag{A2}$$

Clearly, $f(q)$ is concave on $[0, p^2/4]$. Also, we have $f(0) \ge 0$ and $f(p^2/4) \ge 0$. Thus, $f(q) \ge 0$ on $[0, p^2/4]$.

We are done.

River Li
  • 37,323