<s>[INST] 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!
 [/INST] Note $2ab\le a^2+b^2$ and
$$\frac{a+b}c = \sqrt{\frac{(a+b)^2}{a^2+b^2}}= \sqrt{1+\frac{2ab}{a^2+b^2} }\le \sqrt{1+1} =\sqrt2
$$
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