# Inequality proof using real numbers

If $x,y,z,w$ are positive real numbers such that $x < y$ and $z < w$, show that $xz < yw$. Show the converse and prove it or provide a counterexample.

So I know that the proof is true, I just have no idea how to prove it. It feels like its a fact to me. Any help will be appreciated!

• Note the converse isn't true. A counter is: $2\cdot 2 < 1\cdot 5$. – Macavity Dec 12 '14 at 5:55

Note that $$xz - yw = x \cdot \left( z - w \right) + w \cdot \left( x - y \right).$$Since $x-y<0$ and $z-w<0$, we have $xz - yw < 0$, so $xz < yw$. $\Box$

• @ZoëSoriano Does this answer your question? – Ahaan S. Rungta Dec 12 '14 at 5:41
• You're welcome; thanks for the accept! – Ahaan S. Rungta Dec 12 '14 at 7:15

HINT:

Use the following $$xz-yw=x(z-w)+w(x-y)$$

or $$xz-yw=z(x-y)+y(z-w)$$

$x<y \implies \dfrac{x}{y}<1$

$z<w \implies \dfrac{z}{w}<1$

$\therefore \dfrac{xz}{yw}<1$

For this I have used the following lemma.

Lemma

$a<1, b<1 \implies ab<1$

Proof

$a<1 \implies \exists \varepsilon_1>0:a=1-\varepsilon_1$

$b<1\implies \exists \varepsilon_2>0:b=1-\varepsilon_2$

$ab=\left(1-\varepsilon_1\right)\left(1-\varepsilon_2\right)=1-\left(\varepsilon_1+\varepsilon_2\right)+\varepsilon_1\varepsilon_2$

$$\boxed{\left(\varepsilon_1+\varepsilon_2\right)>\varepsilon_1\varepsilon_2 \impliedby \varepsilon_1>\varepsilon_2\left(\varepsilon_1-1\right) \impliedby \varepsilon_1>0}$$