prove this inequality $(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3\le27 (xyz)^2$

Question

Let $x,y,z>0$,show that $$(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3\le27 (xyz)^2$$

I have prove this inequality $$(x+y-z)(y+z-x)(x+y-z)\le xyz$$ because it is three schur inequality $$\Longleftrightarrow x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(z+x)$$ how to solve this inequiality $xyz ⩾ (x+y-z)(y+z-x)(z+x-y)$ But I can't prove this inequality to prove $(1)$

Answer 1

It's enough to prove our inequality for $\prod\limits_{cyc}(x+y-z)\geq0.$

Now, if $x+y-z\leq0$ and $x+z-y\leq0$ then $x+y-z+x+z-y\leq0$ or $2x\leq0$,

which is a contradiction.

Thus, we can assume that $x+y-z>0$, $x+z-y>0$ and $y+z-x>0.$

Now, let $x+y-z=c$, $x+z-y=b$ and $y+z-x=a$.

Thus, we need to prove prove that $$27(a+b)^2(a+c)^2(b+c)^2\geq64abc(a+b+c)^3$$ and since $$(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)$$ it's just $$\sum_{cyc}c(a-b)^2\geq0,$$ it's enough to prove that $$3(ab+ac+bc)(a+b)(a+c)(b+c)\geq8abc(a+b+c)^2$$ or $$\sum_{cyc}(3a^3b^2+3a^3c^2-2a^3bc-4a^2b^2c)\geq0,$$ which is true by Muirhead.

A bit of easier way it's to use $$(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)$$ for the all term $\prod\limits_{cyc}(a+b)^2.$

If so, it's enough to prove that $$(ab+ac+bc)^2\geq3abc(a+b+c),$$ which is obvious.

The following much more stronger inequality is also true.

Let $x$, $y$ and $z$ be real numbers. Prove that: $$4x^2y^2z^2\geq(x+y-z)(x+z-y)(y+z-x)(x^3+y^3+z^3+xyz).$$

Answer 2

We can assume that $x,y,z$ are sides of a triangle because otherwise the LHS is negative.

By Heron's formula we have

$$A^2 = s(s-x)(s-y)(s-z) = \frac1{16}(x+y+z)(x+y-z)(y+z-x)(x+z-y)$$

where $A$ is the area and $s$ is the semiperimeter.

Now recall this inequality:

$$4\sqrt{3}A \le \frac{9xyz}{x+y+z}$$

or $$A^2 \le \frac{27}{16} \frac{(xyz)^2}{(x+y+z)^2}$$

It is listed on wikipedia and for a proof see here.

We get

\begin{align} (x+y-z)(y+z-x)(x+z-y)(x+y+z)^3 = 16{A^2}(x+y+z)^2 \le {27(xyz)^2} \end{align}

Answer 3

$$\color{brown}{\textbf{Final edition (13.08.18)}}$$ $\color{green}{\textbf{Task transformations.}}$

The problem is homogeneous with respect to unknowns $x,y,z.$

Let WLOG $$x+y+z=6,\tag1$$ Then the equivalent inequality is $$x^2y^2(6-x-y)^2\ge64(3-x)(3-y)(x+y-3),\tag2$$ where $$(x,y,6-x-y)\in(0,6),$$ or $$((3-y)+(x+y-3),(3-x)+(x+y-3),(3-x)+(3-y))\in(0,6).\tag3$$

Inequality $(2)$ is satisfied if one or three of the factors $(3-x),(3-x),(x+y-3)$ are non-positive.
On the other hand, conditions $(3)$ is not satisfied if two of this tactors are non-positive.

Thus, it remains to prove the inequality $(2)$ in the case $$(x,y,x+y-3)\in(0,3)^3.\tag4$$

Denote $$t= xy,\tag5$$ then $$(3-x)(3-y)= xy - 3(x+y) +9 = t-3(6-z)+9 = t+3z-9,$$ $$t=xy\le\left(\frac{x+y}2\right)^2=\frac{(6-z)^2}4,$$ and the stronger problem is to prove the inequality $$z^2t^2\ge 64(t+3z-9)(3-z),$$ or $$f(z,t)=z^2t^2+64(z-3)t + 192(z-3)^2\ge0,\tag6$$ under the conditions $$z\in(0,3),\quad t\in\left(0,\frac{(6-z)^2}4\right].\tag7$$

$\color{green}{\textbf{Proof.}}$

The least value of $f(z,p)$ can be achieved only in the stationary points or in the bounds.

The stationary points of $f(z,p)$ can be found from the system $f'z=f'_t = 0,$ or \begin{cases} zt^2+32t + 192(z-3) = 0\\ z^2t+32(z-3) = 0. \end{cases} Taking in account $(7),$ one can obtain $$\begin{cases} (zt+32-6z^2)t = 0\\ z^2t+32(z-3) = 0, \end{cases}\rightarrow \begin{cases} zt=6z^2-32\\ z(6z^2-32)+32(z-3) = 0, \end{cases}\rightarrow \begin{cases} z_s=2\sqrt[3]2\\ t_s=12\sqrt[3]2-8\sqrt[3]4 \end{cases}\\[8pt] f(z_s,t_s) = 64(11-54\sqrt[3]2+36\sqrt[3]4)\approx7.085>0. $$

Let us check the bounds.

Case $\mathbf{z\to 0,\quad t\in(0,9).}$ $$f(0, t) = 192(9 - t)>0.$$

Case $\mathbf{z\to 3, quad t > 0.}$ $$f(3, t) = 3t^2 >0.$$

Case $\mathbf{t=\frac{(6-z)^2}4,\quad z\in(0,3).}$ $$f\left(z, \frac{(6-z)^2}4\right) = \frac1{16}z^2(6-z)^4+16(z-3)(6-z)^2 + 192(z-3)^2 = \frac1{16}z^2(z-2)^2(z^2-20z+132)\ge0$$ (see also Wolfram Alpha).

$\textbf{Proved.}$