sqing wrote: Let $a, b, c $ be the side lengths of a right angled triangle with c > a, b. Show that $$3<\frac{c^3-a^3-b^3}{c(c-a)(c-b)}\leq \sqrt{2}+2.$$ Done!

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sqing wrote: Let $a, b, c $ be the side lengths of a right angled triangle with c > a, b. Show that $$3<\frac{c^3-a^3-b^3}{c(c-a)(c-b)}\leq \sqrt{2}+2.$$ $$\frac{c^3-a^3-b^3}{c(c-a)(c-b)}=\frac{2c+a+b}{c},$$$$3<\frac{c^3-a^3-b^3}{c(c-a)(c-b)}\leq \sqrt{2}+2\iff c<a+b\leq \sqrt{2}c.$$

$c^3 - a^3 - b^3 = c(a^2 + b^2) - a^3 - b^3 = a^2(c-a) + b^2(c-b) = (c-b)(c+b)(c-a) + (c-a)(c+a)(c-b) = ((c+a)+(c+b))((c-a)(c-b))$. $\frac{c^3-a^3-b^3}{c(c-a)(c-b)} = \frac{(c+a)+(c+b)}{c} > 3$ $\frac{(c+a)+(c+b)}{c} = 2 + \frac{a+b}{c} = 2 + \frac{\sqrt{c^2 + 2ab}}{c}$ it's maximum when $ab$ is maximum which is when $a = b = \frac{c}{\sqrt{2}}$ which then $\frac{a+b}{c} = \sqrt{2}$ so $\frac{(c+a)+(c+b)}{c} \le 2 + \sqrt{2}$.