mjhjmtu

I'm not sure if this counts as a "simple answer", but let $x$ be the length of one leg of a right triangle of area $7$; then the other leg is $frac14x$, and the hypotenuse is $sqrtx^2 + left(frac14xright)^2$, so the perimeter is given by the function
$$P(x) = x + frac14x + sqrtx^2 + left(frac14xright)^2$$

Asymptotically, we have $lim_xto 0 P(x) = +infty$ and $lim_xtoinfty P(x) = +infty$, and intuitively it seems clear that the absolute minimum occurs when the two legs of the triangle are equal in length, i.e. when $x=sqrt14$. At this value, we have
$$P(sqrt14) = 2sqrt14 +2sqrt7$$
This is the smallest possible perimeter for a right triangle of area $7$. It remains only to convince yourself that this number is greater than $12$.

If the perimeter is $12$, the triangle of maximum area is the equilateral triangle of side $4$ which has area $2times 2sqrt 3lt 7$ (square both sides $48lt 49$)

Now it is a question of how you show that the triangle of maximum area is equilateral. First fix a base AB. The third point C lies on an ellipse, and symmetry/convexity show that the maximum area is if the two remaining sides AC and BC are equal. So if the sides are not equal, you will get a greater area by averaging the shortest and longest. For maximum area (no further gains) all sides must be equal - hence equilateral.

In summary, a more formal way of showing that the maximal area of a triangle of fixed perimeter is reached when the triangle is equilateral is to go via Heron's formula (the area of a triangle as a function of its sides) and AM/GM. With the perimeter $2s=a+b+c$ fixed, note that $(-a+b+c)+(a-b+c)+(a+b-c)=2s$ and the product of the three terms in brackets is therefore greatest when they are equal.

Heron's formula is area =$sqrt s(s-a)(s-b)(s-c)$

Note: This answer is to an earlier version of the question which asked for perimeter $14$.

What about $$
beginalign
a&=4+sqrt2\
b&=4-sqrt2\
c&=6
endalign
$$

If the area was $7$ and the perimeter was $12$, the radius of the inscribed circle would be $14/12$ (exercise: why?). The circle is on the picture below.

On the other hand you can create any possible right triangle with perimeter $12$ by picking a point $H$ on the arc between $U$ and $V$ ($T$ is the center) and plotting a hypotenuse tangent to the arc at $H$ (exercise: why is the perimeter independent of $H$, if $H$ is on the arc?).

Imagine you move $H$ on the arc, generating all possible triangles. None of them would have the given circle as its inscribed circle. The radius of the circle is just to large (or the radius of the arc is too small).

Note: This answer is to an earlier version of the question which asked for perimeter $14$.

There is indeed such a triangle.

The isosceles right triangle with legs $14/(2+sqrt2)$ has perimeter $14$ and area

$$
frac12left(frac142+sqrt2right)^2=147-98sqrt2approx8.4;.
$$

On the other hand, a right triangle with perimeter $14$ can be made to have arbitrarily small area by making one of the legs arbitrarily small. The right triangles with perimeter $14$ can be parametrized by the length of the shorter leg. The area is a continuous function of this parameter. By the intermediate value theorem, there is a right triangle with perimeter $14$ with some intermediate length of the shorter leg that has area $7$.

$$a+b=12-c$$
$$(a^2+b^2)+2ab=(a+b)^2=c^2-24c+144$$
$$c^2+28=c^2-24c+144$$
$$c=frac296$$
$$a+b=frac436implies (a,b,c)=(k,frac436-k,frac296)$$
Since $c>a, c>b$, $$kin(0, frac296)$$

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

beginalign
x^3-2rho,x^2+(rho^2+r^2+4,r,R),x-4,rrho,R
&=0
tag1label1
.
endalign

For the right triangle
the equation eqref1
can be simplified using a well-known condition

beginalign
R&=tfrac12(rho-r)
tag2label2
endalign
as follows:

beginalign
(x+r-rho)(x^2-(r+rho)x+2rrho)
&=0
tag3label3
.
endalign

The first root, provided by the linear term in eqref3
is the size of hypotenuse,

beginalign
c&=rho-r
,
endalign

the other two sides must be

beginalign
a,b&=tfrac12left(r+rhopmsqrtr^2-6rrho+rho^2right)
.
endalign

Or, in terms of the area $S$ and perimeter $p$,

beginalign
c&=tfrac12p-tfrac2Sp
,\
a,b&=
frac4S+p^2pmsqrtp^4-24Sp^2+16S^24p
tag4label4
.
endalign

So, given $S=7$, $p=12$ we get

beginalign
c&=tfrac296
,\
a,b&=tfrac112(43pmsqrt167,i)
,
endalign

thus, the right triangle with declared properties is impossible.

Using eqref4 it is trivial to find out that
the right triangle with $S=7$, $p=14$
has $c=6$, $a,b=4pmsqrt2$.

And as a bonus, the right triangle with $p=13$
is also valid and has side lengths

beginalign
c&=tfrac14126approx 5.423
,\
a&=tfrac152(197-sqrt953)approx 3.195
,\
b&=tfrac152(197+sqrt953)approx 4.382
endalign

and is just slightly bigger than
the famous $3-4-5$ right triangle.

Let me fill in a missing detail from the answer of @mweiss, namely, that a non-isosceles right triangle of perimeter $17$ does not have minimal area among all right triangles of perimeter $17$. And the specific value of perimeter does not matter for this purpose, so I'll do it for perimeter $p$.

Letting the legs of the triangle be $a,b$, from the Pythagorean theorem we get
$$a^2 + b^2 = (p-a-b)^2
$$
and simplifying we get
$$2p a + 2p b - 2ab = (2p)^2
$$
Differentiating implicitly with respect to $a$ we get
$$2p + 2p fracdbda - 2b - 2 a fracdbda = 0
$$
and so
$$fracdbda = -fracp-bp-a qquad(*)
$$
Since the area is $A=fracab2$ we also get
$$2pa + 2pb - 4A = (2p)^2
$$
Differentiating with respect to $a$ we get
$$2p + 2p fracdbda - 4 fracdAda = 0
$$
Assuming $A$ is a minimum we get
$$2p + 2p fracdbda = 0
$$
and so $fracdbda=-1$. Combining this with $(*)$ and simplifying we get
$$a=b
$$

The triangle is two adjacent sides and one diagonal of a rectangle. As you pointed out, the product of the two legs is 14. This is the rectangle's area.

Making the rectangle a square minimises the sum of the triangle's legs and also minimises the rectangle's diagonal (the triangle's hypotenuse). This thus minimises the triangle's perimeter. Each leg is $sqrt14$ and the hypotenuse is $sqrt28$. This gives a perimeter $>12.7$. Thus no such triangle with perimeter 12 exists.

WLOG let $c>a>b$
$$14=abspace (0)$$
$$a+b+c=12space (1)$$
$$3c>12space (2)$$
$$c>4space (3)$$
$$a+b+c>4+a+bspace (4)$$
$$12>4+a+bspace (5)$$
$$8>a+bspace (6)$$
$$4>fraca+b2space (7)$$
$$16>left(fraca+b2right)^2space (8)$$
$$16>fraca^2+2ab+b^24space (9)$$
$$2>fraca^2-2ab+b^24space(10)=(9)-(0)$$
$$2>left(fraca-b2right)^2space (11)$$
$$sqrt2>fraca-b2space (12)$$
$$4+sqrt2>aspace (13)= (7)+(12) $$
$$4-sqrt2>bspace (14)=(7)-(12)$$
$$14>abspacespace (15)=(13)cdot (14)$$
(15) contradicts (0) QED