MHB Right-Angled Triangle Inequality

[anemone]

Show that if $a,\,b$ and $c$ are the lengths of the sides of a right triangle with hypotenuse $c$, then

$$\frac{(c − a)(c − b)}{(c + a)(c + b)}\le 17 − 12\sqrt{2}$$

 

[kaliprasad]

Show that if $a,\,b$ and $c$ are the lengths of the sides of a right triangle with hypotenuse $c$, then

$$\frac{(c − a)(c − b)}{(c + a)(c + b)}\le 17 − 12\sqrt{2}$$

Spoiler

From law of symmetry we shall have this shall have extremum at a = b giving $c= \sqrt{2}a$ and
$LHS = \frac{(c − a)(c − b)}{(c + a)(c + b)} = \frac{(\sqrt{2}-1)^2}{(\sqrt{2}+1)^2}= (\sqrt{2}-1)^4 = 17 - 12\sqrt{2}$
taking another point (a = 0, c=b) we get LHS = 0
hence $\frac{(c − a)(c − b)}{(c + a)(c + b)}\le 17 − 12\sqrt{2}$

 

[Albert1]

Spoiler

From law of symmetry we shall have this shall have extremum at a = b giving $c= \sqrt{2}a$ and
$LHS = \frac{(c − a)(c − b)}{(c + a)(c + b)} = \frac{(\sqrt{2}-1)^2}{(\sqrt{2}+1)^2}= (\sqrt{2}-1)^4 = 17 - 12\sqrt{2}$
taking another point (a = 0, c=b) we get LHS = 0
hence $\frac{(c − a)(c − b)}{(c + a)(c + b)}\le 17 − 12\sqrt{2}$

Spoiler

$a,b,\,\, and\,, c$ are the lengths of the sides of a right triangle
$a$ cannot be $"0"$
and more :
$c$ is hypotenuse ,we have $b<c$

 

[kaliprasad]

Spoiler

$a,b,\,\, and\,, c$ are the lengths of the sides of a right triangle
$a$ cannot be $"0"$
and more :
$c$ is hypotenuse ,we have $b<c$

Spoiler

What you are telling is right. I took the limiting case

 

[anemone]

My solution:

Spoiler

We have $c^2−a^2=b^2 \implies c−a=\dfrac{b^2}{c+a}$.

By the similar token, we also have $c−b=\dfrac{a^2}{c+b}$, if we're going to replace these two into the original LHS of the inequality, we get:$$\begin{align*}\frac{(c − a)(c − b)}{(c + a)(c + b)}&=\frac{(b^2)(a^2)}{(c + a)^2(c + b)^2}\\&=\left(\frac{ab}{(c + a)(c + b)}\right)^2\\&=\left(\frac{ab}{c^2+c(a+b)+ab}\right)^2\\&=\left(\frac{1}{\dfrac{c^2}{ab}+\dfrac{c(a+b)}{ab}+1}\right)^2\\&\le \left(\frac{1}{\dfrac{c^2}{ab}+\dfrac{c(2\sqrt{ab})}{ab}+1}\right)^2\,\,\text{(by the AM-GM inequality)}\\& = \left(\frac{1}{\left(\dfrac{c}{\sqrt{ab}}\right)^2+\dfrac{2c}{\sqrt{ab}}+1}\right)^2\text{but from $c^2=a^2+b^2\ge 2ab$, we get $\dfrac{c}{\sqrt{ab}}\ge \sqrt{2}$}\\& \le \left(\frac{1}{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}\right)^2\\&= \left(3-2\sqrt{2}\right)^2\\&= 17-12\sqrt{2}\,\,\,\text{Q.E.D.}\end{align*}$$
Equality occurs when $a=b$.