How to prove this inequality? $ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$

let $a,b,c,d\ge 0$,and $a^2+b^2+c^2+d^2=3$,prove that

$ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$

I find this inequality are same as Crux 3059 Problem.

• You may find some solution attempts for problem 3059 online. E.g. see here or here. Your problem is only slightly different, so the same solution techniques may work too. – user1551 Mar 15 '13 at 7:09
• Try optimizing $a+b+c+d+2abcd-ab-ac-ad-bc-bd-cd$ over the region you have described (which is a section of a 3-sphere in $\mathbb{R}^4$.) If you show this quantity's smallest value is at least $0$, then your inequality follows. Lagrange multipliers seem like a good tool for this quantity and this constraint. – alex.jordan Mar 15 '13 at 7:13
• I given this Crux3059 problem methos, But I failure it. – math110 Mar 15 '13 at 7:17
• I read the linked above solution of Crux3059 problem, and I have doubts in its usability for present problem. This solution begins with the consecutive replacing of the pair of variables by square-root of the mean of their squares. But we may not such replace the pair of variables, because we have an additional summand $(a+b+c+d)$. – Alex Ravsky May 15 '13 at 7:12
• @CommonerG The technique in the link does not work in this problem. – Shuhao Cao May 15 '13 at 17:07

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.

1. $w^3=0$.

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.

1. $y=z=1$.

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!

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.

1. 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 $f(a,b,c,d)\ge 0$.

2. $\operatorname{rank}J(x)<2$, 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 $f(a,b,c,d)=4\sqrt{3}-27/4>0$.

3. 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

Your constraints are $a,b,c,d > 0$ and $a^2+b^2+c^2+d^2 = 3$, by doing a change of variables, $x^2 = a^2/3, y^2 = b^2/3, z^2 = c^2/3, w^2 = d^2/3$. Then your constraints change to

$$x,y,z,w > 0, x^2+y^2+z^2+w^2 =1$$

and the problem is equivalent to proving $xy +xz +xw +yz+yw+zw \leq \sqrt{3}(x+y+z+w) + \frac{2}{3}xyzw$. Since $x^2+y^2+z^2+w^2 =1$ that first implies that $x,y,z,w \leq 1$ and in particular $x^2+y^2+z^2+w^2 \leq x+y+z+w$. Doing your standard expansion,

$$(x+y+z+w)^2 = x^2+y^2+z^2+w^2 + 2(xy+xz+xw+yz+yw+zw)$$

The right hand side is bounded above by $x^2+y^2+z^2+w^2 + (x^2+z^2)+(x^2+w^2)+(y^2+z^2) +(y^2 + w^2) + 2(xy + wz) = 3(x^2+y^2+z^2+w^2) + 2(xy+wz)$

Balancing this with the RHS of the display we get

$$2(xy+xz+xw+yz+yw+zw) \leq 2(x^2+y^2+z^2+w^2) + 2(xy+wz)$$

Then we have that $xy \leq (x^2+y^2)/2$ and $wz \leq (w^2+z^2)/2$ and placing this in we have

$$xy+xz+xw+yz+yw+zw \leq 1.5(x^2+y^2+z^2+w^2) \leq \sqrt{3}(x+y+z+w) + \frac{2}{3}xyzw$$

since $\frac{2}{3}xyzw \geq 0$, $1.5 <\sqrt{3}$ and $x^2+y^2+z^2+w^2 \leq x+y+z+w$.

• The change of variables should give $3(xy+xz+xw+yz+yw+zw) \leq \sqrt{3}(x+y+z+w)+18xyzw$, so your proof doesn't work. – Ivan Loh May 15 '13 at 4:02
• thankyou for that, what a silly mistake to make, i will see if i can adjust it, if not, delete my suggestion – user61038 May 15 '13 at 4:17

Let $a=\sqrt 3\cos x\cos y,b=\sqrt 3\cos x\sin y,c=\sqrt 3\sin x\cos y,d=\sqrt 3\sin x\sin y, \pi/2>x,y>0$.

• It is very difficult to read your answer as written and it helps to format them using MathJax (see FAQ). Regards – Amzoti Mar 21 '13 at 20:23
• Then $ad=bc$. :-) – Alex Ravsky May 12 '13 at 19:14