Prove that $(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$

Question

Let $a,b,c,d>0$. Prove that $$(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$$

I don't know how to begin to solve this problem

Answer 1

Let $a=\min\{a,b,c,d\}$, $b=a+u$, $c=a+v$ and $d=a+w$.

Hence, $(a+b)(b+c)(c+d)(d+a)-\prod\limits_{cyc}(a+b+c-d)=$ $$4(u^2+v^2+w^2-uv-vw)a^2+(4(u^3+v^3+w^3)-2u^2v-2v^2u-2v^2w-2w^2v)a+$$ $$+\sum\limits_{cyc}(u^4-2u^2v^2+u^2vw)+u^2w^2$$ Since $\sum\limits_{cyc}(2u^3-u^2v-u^2w)=\sum\limits_{cyc}(u+v)(u-v)^2\geq0$,

it remains to prove that $\sum\limits_{cyc}(u^4-2u^2v^2+u^2vw)\geq0$, which is Schur:

$$\sum\limits_{cyc}(u^4-2u^2v^2+u^2vw)=\sum\limits_{cyc}(u^4-u^3v-u^3w+u^2vw)+\sum\limits_{cyc}uv(u-v)^2\geq0$$ Done!

Answer 2

Also we can use the following reasoning.

Since $\prod\limits_{cyc}(a+b)-(a+b+c+d)(abc+abd+acd+bcd)=(ac-bd)^2\geq0$,

it remains to prove that $$(a+b+c+d)(abc+abd+acd+bcd)\geq\prod\limits_{cyc}(a+b+c-d)$$ which is also true by BW.

Answer 3

We can assume that $a+b+c+d=2$. Then the inequality becomes $$ (1-d)(1-c)(1-b)(1-a)\le\left(1-\tfrac{c+d}2\right)\left(1-\tfrac{a+d}2\right)\left(1-\tfrac{a+b}2\right)\left(1-\tfrac{c+b}2\right)\tag{1} $$ If any of $a$, $b$, $c$, or $d$ is greater than $1$, then the left side is negative and the inequality is trivial. So we can assume $0\le a,b,c,d\le1$.

Subsituting $a\mapsto1-a$, $b\mapsto1-b$, $c\mapsto1-c$, and $d\mapsto1-d$ shows that the inequality is equivalent to $$ \begin{align} abcd &\le\left(\frac{c+d}2\right)\left(\frac{a+d}2\right)\left(\frac{a+b}2\right)\left(\frac{c+b}2\right)\\ &=\left(\frac{ac-bd}4\right)^2+\left(\frac1a+\frac1b+\frac1c+\frac1d\right)\frac{abcd}8\tag{2} \end{align} $$ Since $\frac1x$ is convex for $x\gt0$, Jensen's Inequality says that $$ \begin{align} \frac14\left(\frac1a+\frac1b+\frac1c+\frac1d\right)&\ge\frac1{\frac14(a+b+c+d)}\\ &=2\tag{3} \end{align} $$ $(3)$ shows that $(2)$ is true, which in turn shows that $(1)$ is true.

Answer 4

The simplest method to solve this question should be the following:

Let

$$b+c+d-a=w\tag{1}$$

$$a+c+d-b=x\tag{2}$$

$$a+b+d-c=y\tag{3}$$

$$a+b+c-d=z\tag{4}$$

So, the given inequality is equivalent to

$$16wxyz \leq (w+x)(x+y)(y+z)(z+w)\tag{*}$$

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It is not hard to see that at most one of $w,x,y,z$ is less than zero.

So, there are two cases to consider:

$(1)$ All of $w,x,y,z \geq 0$.

The result is true by AM-GM Inequality

Equality holds when $a=b=c=d$

$(2)$ One of $w,x,y,z$ , say $w<0$ and $x,y,z \geq 0$

$\text{L.H.S } < 0$, while $\text{R.H.S } > 0$

So, inequality holds.