A right triangle has legs $a$ and $b$ and hypotenuse $c.$ Find the largest possible value of $\frac{a + b}{c}$. 
A right triangle has legs $a$ and $b$ and hypotenuse $c.$ Find the largest possible value of
  $$\frac{a + b}{c}.$$


I used the QM-AM inequality. For a set of numbers $\{a_1, a_2, \dots, a_n\}$ this inequality is,
$$\sqrt{\frac{a_1^2 + a_2^2+\dots+a_n^2}{n}}\geq\frac{a_1 + a_2 + \dots + a_n}{n}.$$
Since $(a,b,c)$ form a right triangle with $c$ being the hypotenuse, we have $a^2 + b^2 = c^2$. Using the QM-AM inequality on $a, b, $ and $c$, we have, 
$$\frac{a+b+c}{3}\leq\sqrt{\frac{a^2+b^2+c^2}{3}}.$$
Multiplying both sides by $9$ and plugging in $a^2 + b^2 =c^2$, we have
$$a+b+c\leq \sqrt{6c^2}.$$
Taking the $c$ out of the square root, we have, 
$$a+b+c\leq c\sqrt6.$$
Subtracting $c$ from both sides, we have, 
$$a+b\leq c\sqrt{6}-c.$$
Factoring $c$ out, we have $$a+b\leq(1-\sqrt{6})\cdot c.$$ I don't know where to go from here. I know for a fact that $1-\sqrt{6}$ isn't the answer, and this is probably because $a, b, c$ have to be positive, but I don't know how to solve for it. Any help? Thanks!

A suggestion, if you may.
In terms of one of the acute angles $\theta$ in that triangle the value to maximize is:
$\sin \theta + \cos \theta =\sqrt{2} \cos \left(\theta - \dfrac{\pi}4 \right)$
https://math.stackexchange.com/a/3646439/719444
And for $0 \le \theta \le \dfrac{\pi}2$, there is a maximum at $\theta=\dfrac{\pi}4$ which gives the same answer in general.
