2

Given that $ x, y, z \in \mathbb{R}^{+}$, prove or disprove the inequality

$$ \dfrac{xy}{z} + \dfrac{yz}{x} + \dfrac{zx}{y} \ge x + y + z $$

I have rearranged the above to:

$$ x^2y(y - z) + y^2z(z - x) + z^2x(x - y) \ge 0 \\ \text{and, } \dfrac{1}{x^2} + \dfrac{1}{y^2} + \dfrac{1}{z^2} \ge \dfrac{1}{xy} + \dfrac{1}{xz} + \dfrac{1}{yz} $$ What now? I thought of making use of the arithmetic and geometric mean properties:

$$ \dfrac{x^2 + y^2 + z^2}{3} \ge \sqrt[3]{(xyz)^2} \\ \text{and, } \dfrac{x + y + z}{3} \ge \sqrt[3]{xyz} $$ but I am not sure how, or whether that'd help me at all.

hjpotter92
  • 3,049
  • You have an answer here http://math.stackexchange.com/questions/576592/show-that-displaystyle-fracxyz-fracxzy-fracyzx-geq-xyz, apply AM-GM three times then sum the inequalities – Hippalectryon Aug 09 '14 at 12:48
  • $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y} \implies \frac{x^2 y^2+x^2 z^2+y^2 z^2}{x y z} \implies \frac{y z}{x}+x (\frac{y}{z}+\frac{z}{y})\\text{ } (x>0 \vee y>0 \vee z>0) \implies \frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y} \geq x+y+z$ – JohnWO Aug 09 '14 at 13:05

3 Answers3

9

Consider this inequality:

$$(a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0$$ Expand and simplify the above expression to get:

$$a^2 + b^2 + c^2 \ge ab + bc + ca$$

Substitute $a = \dfrac{1}{x}, b = \dfrac{1}{y}$ and $c = \dfrac{1}{z}$:

$$\dfrac{1}{x^2} + \dfrac{1}{y^2} + \dfrac{1}{z^2} \ge \dfrac{1}{xy} + \dfrac{1}{yz} + \dfrac{1}{zx}$$

Multiply both sides with $xyz$ to get

$$\dfrac{xy}{z} + \dfrac{yz}{x} + \dfrac{zx}{y} \ge x + y + z$$

lakshayg
  • 482
2

Holder's inequality:

$$u\cdot v \leq |u||v|$$

for any vectors $u,v$.

Let $\mathbf u=(1/x,1/y,1/z)$ and $\mathbf v=(xz,xy,yz)$. Then show $|v| = xyz |u|$ and thus $$|\mathbf u||\mathbf v| = xyz\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)= \dfrac{xy}{z} + \dfrac{yz}{x} + \dfrac{zx}{y} $$ and: $$\mathbf u\cdot\mathbf v = x+y+z $$

Thomas Andrews
  • 177,126
2

Use the rearrangement inequality:

$$xy={xyz\over z},$$ so WLOG(without loss of generality), if $0<x\le y\le z$, $$\begin{gather}\frac1x\ge\frac1y\ge\frac1z,\\ yz\ge zx\ge xy.\end{gather}$$ Then we have the result, since LHS of the given inequality is the maximum value when rearrange.

Jaehyeon Seo
  • 1,873