How to prove this inequality? $ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$
It seems the following.
We can easily transform the inequality to the form $(a+b+c+d-1)^2\le 4abcd+4$.
Now we estimate the infimum of a function $f(a,b,c,d)=4abcd+4-(a+b+c+d-1)^2$ on the set $X=\{(a,b,c,d):a,b,c,d\ge 0$ and $a^2+b^2+c^2+d^2=3\}.$ Since the set $X$ is compact, the function $f$ attains its minimum on $X$ in some point $x=(a,b,c,d)\in X$. Suppose that $a+b+c+d=M$. Then the point $x$ is also a point of an conditional extremum of the function $f_1(a,b,c,d)=abcd$ with the constraints $g_1(a,b,c,d)=a^2+b^2+c^2+d^2-3=0$ and $g_2(a,b,c,d)=a+b+c+d–M=0$. The Lagrange Theorem implies that one of the following cases hold.
One of the numbers $a,b,c,d$ is equal to $0$. Then $0\le a+b+c+d\le 3\sqrt{\frac{a^2+b^2+c^2+d^2}3}=3$ and $(a+b+c+d-1)^2\le 4$ and $f(a,b,c,d)\ge 0$.
$\operatorname{rank}J(x)<2$, where $J$ is the Jacoby matrix $||\frac{\partial g_i(x)}{\partial x_j}||$. Since $J(a,b,c,d)=\left(\begin{matrix}2a & 2b & 2c & 2d\\ 1 & 1 & 1 &1\end{matrix}\right)$, we see that $a=b=c=d$. Then $a=b=c=d=\sqrt{3}/2$ and $f(a,b,c,d)=4\sqrt{3}-27/4>0$.
There exist numbers $\lambda_1$ and $\lambda_2$ such that $(f_1+\lambda_1g_1+\lambda_2g_2)’(a,b,c,d)=0$. Then we have a system of equations.
$\cases{ bcd+2\lambda_1a+\lambda_2=0\\ acd+2\lambda_1b+\lambda_2=0\\ abd+2\lambda_1c+\lambda_2=0\\ abc+2\lambda_1d+\lambda_2=0\\ }$
Hence all numbers $a,b,c,d$ are the roots of the square equation $abcd+2\lambda_1t^2+\lambda_2t=0$. Since the inequality is symmetric, without loss of generality we may assume that one of the following cases hold:
3.1. $a=b$, $c=d$. Then we have $a^2+c^2=3/2$ and $f(a,c)=4a^2c^2+4-(2a+2c-1)^2$. To find the minimal value of $f$, we again use Lagrange Theorem. The cases $a=0$ or $c=0$ were considered above. The Jacoby matrix in this case is equal to $(2a$ $2c)$ and have nonzero rank, provided $a\not=0$ or $c\not=0$. So it rests to consider the case when there exists a number $\lambda$ such that
$(f+\lambda(a^2+c^2))’(a,c)=0$. Then we have a system of equations.
$$\cases{ 8ac^2–4(2a+2c-1)+2\lambda a=0\\ 8a^2c–4(2a+2c-1)+2\lambda c=0\\ }$$
If $a\not=0$ and $c\not=0$, then we obtain that $(2ac^2+2a^2c-2a-2c+1)(c-a)=0.$
Since $a=c$ implies $a=b=c=d$ and this case was considered above, without loss of generality we may assume that $2ac^2+2a^2c-2a-2c+1=0 $. Put $t=a+c$. Then $2ac=t^2-3/2$ and we have $0=2t^3-7t+2=(t+2)(2t^2-4t+1)$. Since $t\ge 0$, we have $t=1\pm\sqrt{2}/2$, $2ac=3/2-t^2=\mp\sqrt{2}$, and $f(a,c)=3\mp 2\sqrt{2}>0$.
3.2. $b=c=d$. Then we have $a^2+3b^2=3$ and $f(a,b)=4ab^3+4-(a+3b-1)^2$. To find the minimal value of $f$, we again use Lagrange Theorem. The cases $a=0$ or $b=0$ were considered above. The Jacoby matrix in this case is equal to $(a$ $3b)$ and have nonzero rank, provided $a\not=0$ or $b\not=0$. So it rests to consider the case when there exists a number $\lambda$ such that
$(f+\lambda(a^2+3b^2))’(a,b)=0$. Then we have a system of equations.
$$\cases{ 4b^3–2(a+3b-1)+2\lambda a=0\\ 12ab^2–4(a+3b-1)+6\lambda b=0\\ }$$
If $a\not=0$ and $b\not=0$, then we obtain that $(2ab^2+2b^3-a-3b+1)(b-a)=0.$
Since $a=b$ implies $a=b=c=d$ and this case was considered above, without loss of generality we may assume that $2ab^2+2b^3-a-3b+1=0$. Then $1-2b^2\not=0$ and $a=(2b^3-3b+1)/(1-2b^2)=\sqrt{3-3b^2}$. If $b=1$, then $a=0$ and this case was considered above. If $b\not=1$, then $(2b^2+2b–1)/(1-2b^2)=\sqrt{(3+3b)/(1-b)}$ and $8b^5+8b^4–10b^3–8b^2+4b+1=0$. Mathcad calculated that this equality has two not real roots, one negative root, and roots $b_1\sim 0.60442635232610682922$ and $b_2\sim 0.82451890455979510644$. In the first case we obtain $f(a,b)\sim 0.409>0$, in the second case we obtain $f(a,b)\sim 0.177>0$.
From the above we can see that the initial inequality becomes an equality iff one of the numbers $a,b,c,d$ if equal to $0$ and the others are equal to $1$.
Lagrange multipliers is OK and not very complex: $f= a+b+c+d+2abcd−ab−ac−ad−bc−bd−cd,g=a^2+b^2+c^2+d^2-3,F=f+\lambda g$
$F_{a}=1+2bcd-(b+c+d)+2a\lambda=0$......<1>
$F_{b}=1+2acd-(a+c+d)+2b\lambda=0$......<2>
$F_{c}=1+2abd-(a+b+d)+2c\lambda=0$......<3>
$F_{d}=1+2abc-(a+b+c)+2d\lambda=0$......<4>
$F_{\lambda}=a^2+b^2+c^2+d^2-3=0$......<5>
<1>-<2>,<2>-3>,<3>-<4>:
$(a-b)(1-2cd+2\lambda)=0$......<6>
$(b-c)(1-2ad+2\lambda)=0$......<7>
$(c-d)(1-2ab+2\lambda)=0$......<8>
WLOG, we only consider 4 cases:
case I: $a-b=0,b-c=0.c-d=0 \to a=b=c=d=\dfrac{\sqrt{3}}{2}$
case II: $a-b=0,b-c=0,1-2ab+2\lambda=0 \to$$ \lambda=\dfrac{2a^2-1}{2}$ if $a^2\not= 0.5$ put into <4>,$d=\dfrac{1+2a^3-3a}{1-2a^2}=\dfrac{(a-1)(2a^2+2a-1)}{1-2a^2}$ put into <5>
$(a-1) \left[ 3(a+1)+(a-1) \left (\dfrac{(2a^2+2a-1)}{1-2a^2} \right)^2 \right]=0 $
$a=1,\to a=b=c=1,d=0$
when $3(a+1)+(a-1) \left (\dfrac{(2a^2+2a-1)}{2a^2-1} \right)^2 =0 \to a<1$
now I will proof for any root $a<1,f>0$, in this case :
$f=3a+d+2a^3d-3a^2-3ad=3a(1-a)+d(2a^3-3a+1)=(1-a)[3a-d(2a^2+2a-1)]$
if $2a^2+2a-1<0$, then $f>0$
if $2a^2+2a-1>0$,$f>0 \iff 3a>d(2a^2+2a-1) \iff$$ 3a^2>(3-3a^2)(2a^2+2a-1)^2 \iff 2a^4+2a^3-2a+1>0 \iff a^2(2a^2+2a-1)+(a-1)^2>0$
which is true!
for $a^2=0.5$, we will have same result.
case III: $a=b,1-2ad+2\lambda=0,1-2ab+2\lambda=0 \to a(b-d)=0$
when $a=b=0,\to c+d=1,c^2+d^2=3 \to d=\dfrac{1+\sqrt{5}}{2} \to c=\dfrac{1-\sqrt{5}}{2} <0$, so there is no solution in this case.
when $b-d=0 \to a=b=d,$ which is same as case II.
case IV: $1-2cd+2\lambda=0,1-2ad+2\lambda=0,1-2ab+2\lambda=0 \to d(a-c)=0 $and $ a(b-d)=0$ again there are 4 sub cases:
A:$a=0,d=0 \to b+c=1$ which is case III.
B:$d=0,b=d $ which is case A
C:$a=c,a=0$,which is case A
D:$a=c,b=d \to 2\lambda=2ab-1 \to 2a^2+2b^2=3,1+2a^2b+2ab^2-2a-2b=0$ WLOG, let $b \ge a \to 1+b-2b^3+2a(b^2-1)=0 \to (1-b)((1+2b+2b^2-2a(1+b))=0 \to b=1,a=\dfrac{\sqrt{2}}{2} $
verify all solutions, when $a=b=c=1,d=0, f=0$, all other cases $ f>0 \to f_{min}=0$.QED
Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2$, $abc+abd+acd+bcd=4w^3$ and $abcd=t^4$.
Since $(a-b)^2(c-d)^2+(a-c)^2(b-d)^2+(a-d)^2(b-c)^2=24(3v^4-4uw^3+t^4)\geq0$
and the condition gives $16u^2-12v^2=3$, it remains to prove that $$4uw^3-3v^4+16u\left(\sqrt{\frac{4u^2-3v^2}{3}}\right)^3\geq4(4u^2-3v^2)v^2$$
We know that $a$, $b$, $c$ and $d$ are non-negative roots of the equation $$X^4-4uX^3+6v^2X^2-4w^3X+t^4=0$$
Hence, by the Rolle's theorem the equation $$(X^4-4uX^3+6v^2X^2-4w^3X+t^4)'=0$$ or $$X^3-3uX^2+3v^2X-w^3=0$$ has three non-negative roots.
Thus, there are non-negatives $x$, $y$ and $z$, for which
$x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
But the inequality $$4uw^3-3v^4+16u\left(\sqrt{\frac{4u^2-3v^2}{3}}\right)^3\geq4(4u^2-3v^2)v^2$$ is a linear inequality of $w^3$, which says that it's enough to prove it
for an extremal value of $w^3$, which happens in the following cases.
- $w^3=0$.
Let $z=0$ and $y=1$.
We need to prove that $$-\frac{x^2}{3}+\frac{16(x+1)}{3}\left(\sqrt{\frac{\frac{4(x+1)^2}{9}-x}{3}}\right)^3\geq\frac{4x}{3}\left(\frac{4(x+1)^2}{9}-x\right)$$ or $$256(x+1)^2(4x^2-x+4)^3\geq243x^2(16x^2+5x+16)^2$$ Let $x^2+1=px$. Hence, $p\geq2$ and we need to prove that $$256(p+2)(4p-1)^3\geq243(16p+5)^2$$ and since $p\geq2$, it's enough to prove that $$256\cdot4\cdot7(4p-1)^2\geq243(16p+5)^2$$ or $$\left(4-\sqrt{\frac{243}{28}}\right)p\geq1+5\sqrt{\frac{243}{256\cdot28}}$$ for which it remains to prove that $$\left(4-\sqrt{\frac{243}{28}}\right)2\geq1+5\sqrt{\frac{243}{256\cdot28}}$$ or $$7\geq\frac{37}{32}\sqrt{\frac{243}{7}}$$ and since $\sqrt{\frac{243}{7}}<6$, we are done in this case.
- $y=z=1$.
We need to prove that $$\frac{4(x+2)x}{3}-\frac{(2x+1)^2}{3}+\frac{16(x+2)}{3}\left(\sqrt{\frac{\frac{4(x+2)^2}{9}-2x-1}{3}}\right)^3\geq$$ $$\geq\frac{4(2x+1)}{3}\left(\frac{4(x+2)^2}{9}-2x-1\right)$$ or $$256(x+2)^2(4x^2-2x+7)^3\geq243(32x^3+12x+37)^2$$ or $$(4x-1)^2(1024x^6+3072x^5-9984x^4+7744x^3+6144x^2-17088x+18565)\geq0$$ which is obviously true.
Done!