In Springer's "Inequalities" book, Exercise $2.5$ is the following:
Let $x,y,z\in\mathbb{R}^{+}\ $ such that $\ x+y+z = 1.\ $ Then:
$$ xy + yz + zx \geq 9xyz.\qquad (1)$$
Proof: Applying AM $\geq$ GM, we get
$$ xy + yz + zx = (xy + yz + zx)(x + y + z) \geq 3 \sqrt[3]{ (xy)(yz)(zx)} \cdot 3 \sqrt[3]{ xyz } = 9xyz. \square $$
Using the square of the HM-GM inequality; that is, for all $a,b\geq 0, $ we have:
$$ \frac{ 4 a^2 b^2 }{ (a+b)^2 } \leq ab\qquad (2), $$
and then comparing $\frac{ 4 a^2 b^2 }{ (a+b)^2 }$ to $ab$ in each term in $(1),$ I propose a stronger inequality than $(1):$
$$ \frac{ x^2 y^2 }{ (x+y)^2 } + \frac{ y^2 z^2 }{ (y+z)^2 } + \frac{ z^2 x^2 }{ (z+x)^2 } \geq \frac{9}{4} xyz\qquad (3) $$
and so if $(3)$ is true, then it can be viewed as an improvement on inequality $(1).$
I have checked the following cases:
- $x=y=z=\frac{1}{3}:\ $ This is the only case where I found equality of $(3)$, although since $(3)$ is stronger than $(1),$ it is not clear that this is the only case where we have equality in $(3);$
- $x=y=\frac{1}{100},\ z=\frac{98}{100};$
- $x=y=\frac{1}{1000},\ z=\frac{998}{1000};$
- $x=y=\frac{1}{4},\ z=\frac{1}{2};$
- $x=\frac{1}{2},\ y=\frac{1}{3},\ z=\frac{1}{2};$
- $x=\frac{1}{3},\ y=\frac{1}{5},\ z=\frac{7}{15};$
and $(3)$ holds in all these cases.
I have been recreationally searching for ways in which the HM-GM inequality might be able to make other inequalities stronger and therefore improve upon the original inequality, but in other inequalities I have tried this for, the proposed improvement using HM-GM has turned out to be false.
$$\sum_{cyc}\frac{ 4x^2 y^2 }{ (x+y)^2 } \ge 9xyz$$ $$\sum_{cyc}\frac{ 4x y z(x+y+z)}{ (x+y)^2z^2 } \ge 9$$ define $a=yz,b=xz,c=xy$ $$\sum_{cyc}\frac{ 4(ab+bc+ac)}{ (a+b)^2 } \ge 9$$
$$\sum_{cyc}\frac{4}{ (a+b)^2 } \ge \frac{9}{ab+bc+ac}$$ which is well known prove thatt $\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$