Prove an inequality using the property of ordering in $\Bbb Z$ > Given that $(ad-bc)bd\le0$ and $(cf-de)df\le0$, prove that $(af-be)bf\le 0$ by manupulating with these properties of order in $\Bbb Z$. • $a ≤ a$ • $a ≤ b$ and $b ≤ c ⇒ a ≤ c$ • $a ≤ b$ and $b ≤ a ⇒ a = b $ • $a ≤ b ⇒ a + c ≤ b + c $ • $0 ≤ a$ and $0 ≤ b ⇒ 0 ≤ ab $ It seems to be harder then I expect. Could someone prove it step- by-step? Thanks a lot!

$(ad-bc)bd \le 0$

$\implies abd^2 - b^2cd \le 0 $

Since $f^2 \ge 0$ , multiplying with it, we get ,

$abd^2f^2 -b^2f^2cd \le 0$ -----A

Also ,

$(cf-de)df \le 0$

$\implies cdf^2 - d^2ef \le 0$

Since $b^2 \ge 0$, multiplying we get,

$cdb^2f^2 - efb^2d^2 \le 0$ ----- B

Now, $A + B$ gives ,

$abd^2f^2 -b^2f^2cd + cdb^2f^2 - efb^2d^2 \le 0 $

$\implies abd^2f^2 - efb^2d^2 \le 0$

$\implies d^2(af-be)bf \le 0 $

$\implies (af-be)bf \le 0 $, since $d^2 \ge 0$