Hint on solving a problem in Euclidean geometry

Question

How do i prove the following problem:

If a quadrilateral has sides of length $a$, $b$, $c$, and $d$, prove that its area $S$ satisfies the following inequality $$4S\leq (a+c)(b+d)$$ with equality holding only for rectangles.

Hint: Twice the area of a triangle is $a b \sin \alpha$, where $\alpha$ is the angle between the sides of lengths $a$, $b$. But how do i use it?

Thanks in advance.

cheesyfluff · Accepted Answer · 2015-11-08T16:02:36.083

0

Hint

The triangle with sides $a$ and $b$ has the area $$A=\frac{1}{2}ab\sin \alpha\leq\frac{1}{2}ab$$ with the expressions equal if and only if $\alpha=\frac{\pi}{2}$. Do the same for each pair of sides and label these areas $A_1$, $A_2$, $A_3$, $A_4$ and note that $$ 2S=A_1+A_2+A_3+A_4$$ and $$ab+ad+cb+cd = (a+c)(b+d)$$

edited Nov 08 '15 at 16:02

answered Nov 08 '15 at 15:29

cheesyfluff

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$S\leq ab$ is not generally true for all quadrilateral – More water plz Nov 08 '15 at 15:32
Right, I just noticed. I'll change it to account for that. – cheesyfluff Nov 08 '15 at 15:32
aren't you suppose to prove 4S≤(a+e)(b+d)? – More water plz Nov 08 '15 at 15:44
He's asking for a hint so I've outlined the general method to prove what he wants. If you can't see the next step, it's to double $2S$ to get the desired inequality. – cheesyfluff Nov 08 '15 at 15:54
$2S\leq \frac{1}{2}(ad+ bc+ ab+ cd)$, then $4S\leq (a+c)(b+d)$ – user Nov 08 '15 at 15:58
I'll change the side $e$ to side $c$ for clarity in both my answer and the question. – cheesyfluff Nov 08 '15 at 16:02
how can i prove that if a triangle has sides of lengths a, b, c, then its area S satisfies the inequality$$4S\sqrt{3} \leq a^{2} + b^{2 }+c^{2}$$ with equality holding only for equilateral triangles. – user Nov 08 '15 at 16:05

Hint on solving a problem in Euclidean geometry

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