Given that $a+b = c+d = 10$ and $|a-b| > |c-d|$, show that $ab<cd$

Question

By substitution

$$ab<cd \iff a(10-a) < c (10-c) $$ $$|a-b|>|c-d| \iff |a-5|>|c-5|$$

By graphing $y(x)=x(10-x)$, this equation is a quadratic that has a turning point at $(x,y)=(5,25)$ and it open downwards. So the closer $x$ is from $5$, the greater the $y$ value. The second inequality is saying $a$ is further away from $5$ than $c$ is so $y(a) < y(c)$.

I am struggling to show this just by using algebraic manipulation. How can this be shown more rigorously?

Answer 1

Hint: $$|a-b|\ge |c-d|$$ $$\to |a-b|^2\ge |c-d|^2$$ $$\to {(a+b)}^2-4ab\ge {(c+d)}^2-4cd$$ Can you end it now?

Answer 2

Alternative approach

$a,b = 5+r, ~5-r$ in some order.

$c,d = 5+s, ~5-s$ in some order.

Since $|a-b| > |c-d|,$ you have that $r > s.$

Edit
See comments following answer. Above should be
$|r| > |s|$, or I should have specified that $r,s$ were positive.

Therefore $(ab) = 5^2 - r^2 < 5^2 - s^2 = (cd)$.