Right-Angled Triangle Inequality

In summary, the Right-Angled Triangle Inequality is a mathematical rule stating that the sum of the lengths of any two sides of a right-angled triangle must be greater than the length of the third side. It is useful in geometry for determining if a triangle is right-angled and for solving problems involving side lengths. The formula for this inequality is c² = a² + b², where c is the length of the hypotenuse and a and b are the lengths of the other two sides. This inequality only applies to right-angled triangles and can be proved using the Pythagorean Theorem.
  • #1
anemone
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Show that if $a,\,b$ and $c$ are the lengths of the sides of a right triangle with hypotenuse $c$, then

\(\displaystyle \frac{(c − a)(c − b)}{(c + a)(c + b)}\le 17 − 12\sqrt{2}\)
 
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  • #2
anemone said:
Show that if $a,\,b$ and $c$ are the lengths of the sides of a right triangle with hypotenuse $c$, then

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

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 $\frac{(c − a)(c − b)}{(c + a)(c + b)}\le 17 − 12\sqrt{2}$
 
  • #3
kaliprasad said:
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 $\frac{(c − a)(c − b)}{(c + a)(c + b)}\le 17 − 12\sqrt{2}$
$a,b,\,\, and\,, c$ are the lengths of the sides of a right triangle
$a$ cannot be $"0"$
and more :
$c$ is hypotenuse ,we have $b<c$
 
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  • #4
Albert said:
$a,b,\,\, and\,, c$ are the lengths of the sides of a right triangle
$a$ cannot be $"0"$
and more :
$c$ is hypotenuse ,we have $b<c$

What you are telling is right. I took the limiting case
 
  • #5
My solution:

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:\(\displaystyle \begin{align*}\frac{(c − a)(c − b)}{(c + a)(c + b)}&=\frac{(b^2)(a^2)}{(c + a)^2(c + b)^2}\\&=\left(\frac{ab}{(c + a)(c + b)}\right)^2\\&=\left(\frac{ab}{c^2+c(a+b)+ab}\right)^2\\&=\left(\frac{1}{\dfrac{c^2}{ab}+\dfrac{c(a+b)}{ab}+1}\right)^2\\&\le \left(\frac{1}{\dfrac{c^2}{ab}+\dfrac{c(2\sqrt{ab})}{ab}+1}\right)^2\,\,\text{(by the AM-GM inequality)}\\& = \left(\frac{1}{\left(\dfrac{c}{\sqrt{ab}}\right)^2+\dfrac{2c}{\sqrt{ab}}+1}\right)^2\text{but from $c^2=a^2+b^2\ge 2ab$, we get $\dfrac{c}{\sqrt{ab}}\ge \sqrt{2}$}\\& \le \left(\frac{1}{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}\right)^2\\&= \left(3-2\sqrt{2}\right)^2\\&= 17-12\sqrt{2}\,\,\,\text{Q.E.D.}\end{align*}\)
Equality occurs when $a=b$.
 

FAQ: Right-Angled Triangle Inequality

What is the Right-Angled Triangle Inequality?

The Right-Angled Triangle Inequality is a mathematical rule that states that the sum of the lengths of any two sides of a right-angled triangle must be greater than the length of the third side.

How is the Right-Angled Triangle Inequality useful in geometry?

The Right-Angled Triangle Inequality is useful in geometry because it helps determine whether a triangle is a right-angled triangle or not. It also helps in solving problems involving the side lengths of a right-angled triangle.

What is the formula for the Right-Angled Triangle Inequality?

The formula for the Right-Angled Triangle Inequality is c² = a² + b², where c is the length of the hypotenuse and a and b are the lengths of the other two sides.

Can the Right-Angled Triangle Inequality be applied to any triangle?

No, the Right-Angled Triangle Inequality only applies to right-angled triangles. For other types of triangles, there are different inequality rules that apply.

How can the Right-Angled Triangle Inequality be proved?

The Right-Angled Triangle Inequality can be proved using the Pythagorean Theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

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