MHB Is it True That for any Pythagorean Triple, $(\frac ca + \frac cb)^2 > 8$?

[magneto1]

Let $(a,b,c)$ be a Pythagorean triple, specifically, a triplet of positive integers with property $a^2 + b^2 = c^2$. Show that $(\frac ca + \frac cb)^2 > 8$.

EDIT: Added a small clarification.

 

[mente oscura]

Let $(a,b,c)$ be a Pythagorean triple, specifically, a triplet of positive integers with property $a^2 + b^2 = c^2$. Show that $(\frac ca + \frac cb)^2 > 8$.

EDIT: Added a small clarification.

Hello.

Spoiler

(\dfrac{c}{a} +\dfrac{c}{b})^2=T

\dfrac{c^2(a+b)^2}{a^2b^2}=T

c^4+2abc^2-a^2b^2T=0

c^2= \dfrac{-2ab \pm \sqrt{4a^2b^2+4a^2b^2T}}{2}

c^2=ab(\sqrt{T+1}-1)

a^2+b^2=ab(\sqrt{T+1}-1)

If T=8:

a^2+b^2=2ab \rightarrow{}(a-b)^2=0

Therefore:

T>8

Regards.

 

[magneto1]

Hello.

Spoiler

(\dfrac{c}{a} +\dfrac{c}{b})^2=T

\dfrac{c^2(a+b)^2}{a^2b^2}=T

c^4+2abc^2-a^2b^2T=0

c^2= \dfrac{-2ab \pm \sqrt{4a^2b^2+4a^2b^2T}}{2}

c^2=ab(\sqrt{T+1}-1)

a^2+b^2=ab(\sqrt{T+1}-1)

If T=8:

a^2+b^2=2ab \rightarrow{}(a-b)^2=0

Therefore:

T>8

Regards.

Two points that I am not clear from in the proof: (1) Why can't $T < 8$? and (2) When $T = 8$, your proof shows that $(a-b)^2 = 0$, which can happen for some $a = b > 0$. Why can't that happen?

 

[mente oscura]

Two points that I am not clear from in the proof: (1) Why can't $T < 8$? and (2) When $T = 8$, your proof shows that $(a-b)^2 = 0$, which can happen for some $a = b > 0$. Why can't that happen?

I have not put it because it seems to me to be trivial:

Spoiler

If a=b:

a^2+b^2=c^2 \rightarrow{}2a^2=c^2 \rightarrow{}c=a \sqrt{2}

Regards.

 

[kaliprasad]

Spoiler

let
$a = c \cos( t)$
$b = c \sin( t)$

$\frac{c}{a}+ \frac{c}{b} = \sec( t )+\csc( t)$
$(\sec (t) +\ cosec( t))^2 = \sec ^ 2 (t) + \csc ^2 (t) + 2 \sec (t) \csc( t)$
=$ \tan ^2 (t )+ 1 + \cot ^2 (t )+ 1 + 2 \sec( t) \csc( t)$
= $(\tan ^2( t )+ \cot ^2( t) ) + 2 + \frac{4} {( 2 \cos( t)\ sin (t ))}$
= $(\tan( t) -\cot ( t) )^2 + 2 + 2 + \frac{4}{\sin (2t) }$
= $(\tan (t) -\cot ( t) )^2 + 4 + \frac{4}{\sin (2t) } $
>= 0 + 4 + 4 or 8
it can be 8 when a= b but a = b means c cannot be integer so 8 is ruled out

so it is > 8

 

[kaliprasad]

I have not put it because it seems to me to be trivial:

Spoiler

If a=b:

a^2+b^2=c^2 \rightarrow{}2a^2=c^2 \rightarrow{}c=a \sqrt{2}

Regards.

Spoiler

a and b cannot be same integers as if a = b then $c = a\sqrt 2$ not an integer

 

[magneto1]

Thanks mente oscura and kaliprasad for participating. Here is my solution. There are actually others for those who want to find them.

Spoiler

Apply AM-GM inequality,
\[
\left( \frac ca + \frac cb \right)^2 = c^2 \left( \frac{a+b}{ab} \right)^2
= \frac{(a^2+b^2)(a+b)^2}{a^2b^2} \geq \frac{2\sqrt{a^2b^2}(2\sqrt{ab})^2}{a^2b^2} = 8,
\]
with equality holds if $a = b$, which cannot be as it would imply $c = a\sqrt{2}$ contradicts with the fact $c$ is an integer.

Bonus Remark: It can be shown that there is no Pythagorean triple and integer $n$ where the equation $\left( \frac ca + \frac cb \right)^2 = n$ holds.

 

[MarkFL]

My solution:

Spoiler

Our objective function is:

$$f(a,b,c)=\left(\frac{c}{a}+\frac{c}{b} \right)^2$$

Using the constraint:

$$a^2+b^2=c^2$$

We then obtain:

$$f(a,b)=\left(a^2+b^2 \right)\left(\frac{1}{a}+\frac{1}{b} \right)^2$$

Observing that we have cyclic symmetry between $a$ and $b$, we know the critical point comes from $a=b$:

$$f(a,a)=8$$

Testing another point $$(a,2a)$$, we find:

$$f(a,2a)=\frac{45}{4}>8$$

And so over the reals, we know:

$$f_{\min}=8$$

Which means over the integers, we must have:

$$8<f$$