<s>[INST] 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. 
 [/INST] Given a general triangle with 
inradius $r$, circumradius $R$
and semiperimeter $\rho=\tfrac12(a+b+c)$, 
the lengths of its sides $a,b,c$ can be found as 
three roots of the cubic equation
\begin{align}
x^3-2\rho\,x^2+(\rho^2+r^2+4\,r\,R)\,x-4\,r\rho\,R
&=0
\tag{1}\label{1}
.
\end{align}
For the right triangle
the equation \eqref{1} 
can be simplified using a well-known condition
\begin{align}
R&=\tfrac12(\rho-r)
\tag{2}\label{2}
\end{align} 
as follows:
\begin{align}
(x+r-\rho)(x^2-(r+\rho)x+2r\rho)
&=0
\tag{3}\label{3}
.
\end{align}
The first root, provided by the linear term in \eqref{3}
is the size of hypotenuse,
\begin{align}
c&=\rho-r
,
\end{align}
the other two sides must be
\begin{align}
a,b&=\tfrac12\left(r+\rho\pm\sqrt{r^2-6r\rho+\rho^2}\right)
.
\end{align}
Or, in terms of the area $S$ and perimeter $p$,
\begin{align}
c&=\tfrac12p-\tfrac{2S}{p}
,\\
a,b&=
\frac{4S+p^2\pm\sqrt{p^4-24Sp^2+16S^2}}{4p}
\tag{4}\label{4}
.
\end{align}
So, given $S=7$, $p=12$ we get
\begin{align}
c&=\tfrac{29}6
,\\
a,b&=\tfrac1{12}(43\pm\sqrt{167}\,i)
,
\end{align}
thus, the right triangle with declared properties is impossible.
Using \eqref{4} it is trivial to find out that
the right triangle with $S=7$, $p=14$ 
has $c=6$, $a,b=4\pm\sqrt2$.
And as a bonus, the right triangle with $p=13$
is also valid and has side lengths
\begin{align}
c&=\tfrac{141}{26}\approx 5.423
,\\
a&=\tfrac1{52}(197-\sqrt{953})\approx 3.195
,\\
b&=\tfrac1{52}(197+\sqrt{953})\approx 4.382
\end{align}
and is just slightly bigger than
the famous $3-4-5$ right triangle.
</s>