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.
Best Answer
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 $\frac{14}x$, and the hypotenuse is $\sqrt{x^2 + \left(\frac{14}x\right)^2}$, so the perimeter is given by the function $$P(x) = x + \frac{14}x + \sqrt{x^2 + \left(\frac{14}x\right)^2}$$
Asymptotically, we have $\lim_{x\to 0} P(x) = +\infty$ and $\lim_{x\to\infty} P(x) = +\infty$, and intuitively it seems clear that the absolute minimum occurs when the two legs of the triangle are equal in length, i.e. when $x=\sqrt{14}$. At this value, we have $$P(\sqrt{14}) = 2\sqrt{14} +2\sqrt{7}$$ 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 $12$.