Minimum Triangle Ratio

$Consider\quad a\quad triangle\quad with\quad c\quad as\quad length\quad of\quad hypotenuse,\quad a\quad as\quad length\quad of\quad opposite\quad side\quad and\quad b\quad as\quad length\quad of\quad adjacent\quad side\quad \\ To\quad find:\quad min\quad of\quad \frac { c }{ a+b } \\ From\quad the\quad considered\quad triangle,\\ \tan { A } =\frac { a }{ b } \Rightarrow \tan { A } +1=\frac { a+b }{ b } \longrightarrow (1)\\ \sec { A } =\frac { c }{ b } \longrightarrow (2)\\ \frac { (2) }{ (1) } =\frac { \sec { A } }{ \tan { A } +1 } =\frac { c }{ a+b } .\\ Hence\quad min\quad of\quad \frac { c }{ a+b } =\quad min\quad of\quad \frac { \sec { A } }{ \tan { A } +1 } =\frac { 1 }{ \sin { A } +\cos { A } } .\\ min\quad of\quad \frac { 1 }{ \sin { A } +\cos { A } } =\quad max\quad of\quad sinA+cosA.\\ max\quad of\quad sinA+cosA\quad occurs\quad at\quad A=\frac { \pi }{ 4 } \quad \\ \Rightarrow \quad max\quad of\quad sinA+cosA\quad =\quad \sqrt { 2 } .\\ \therefore \quad min\quad of\quad \quad \frac { 1 }{ \sin { A } +\cos { A } } =\frac { 1 }{ \sqrt { 2 } } ≈ 0.7071=min\quad of\quad \frac { c }{ a+b } .$

Let the two short sides be $x$ and $y$

So, we need to find the minimum of :

$\frac{\sqrt{x^2+y^2}}{x+y}$

By Titu's Lemma it follows that :

$\frac{x^2}{1}+\frac{y^2}{1} \ge \frac{(x+y)^2}{2}$

So,

$\frac{\sqrt{x^2+y^2}}{x+y} \ge \frac{1}{\sqrt{2}}$

Hence , the required ratio is $\frac{1}{√2}$ .

Find min of $f(a,b) = \frac{\sqrt{a^2 + b^2}}{a +b}$ being $a$ , $b$ opposite sides of hypotenuse. $\dfrac{df (a,b)}{da} = \frac{\frac{a}{\sqrt{a^2 +b^2}} \cdot (a + b) - \sqrt{a^2 +b^2}}{(a + b)^2} = \frac{ab - b^2}{\sqrt{a^2 +b^2} \cdot (a + b)^2} = 0 \Rightarrow a = b$ $\dfrac{df (a,b)}{db} = \frac{\frac{b}{\sqrt{a^2 +b^2}} \cdot (a + b) - \sqrt{a^2 +b^2}}{(a + b)^2} = 0$ Hence, min of $f(a,b) = \frac{\sqrt{a^2 + b^2}}{a +b}$ is $f(a,a) = \frac{\sqrt{2} a}{2a} = \frac{\sqrt{2}}{ 2} \approx 0.7071$ , because, f is a infinite diferentiable function in the open set $A = \{ (a,b) \in \mathbb{R}^2 \text{ / } a > 0, b > 0\}$ and each absolute extremum is a relative extremum $f(a,a) < f(3,4)$