Does there exist a right triangle with area 7 and perimeter 12? This question is really trivial. I can prove that there is no right triangle with area 7 and perimeter 12, but what I do is solve the following system: if $a$, $b$ and $c$ are, respectively, the two legs and hypotenuse of such a triangle, then
$$a^2 + b^2 = c^2,$$
$$a + b +c = 12,$$
$$ab = 14$$
It is easy (although a bit boring and long) to see that there are no real solutions to this system. 
But I feel that there is a simple answer to this question - perhaps using the triangle inequality - but I just cannot see it. 

$$a+b=12-c$$
$$(a^2+b^2)+2ab=(a+b)^2=c^2-24c+144$$
$$c^2+28=c^2-24c+144$$
$$c=\frac{29}6$$
$$a+b=\frac{43}6\implies (a,b,c)=(k,\frac{43}6-k,\frac{29}6)$$
Since $c>a, c>b$, $$k\in(0, \frac{29}6)$$
