Proving $\sqrt{2}+\sqrt{3} \gt \pi$: A Mathematical Challenge

[anemone]

Prove $\sqrt{2}+\sqrt{3}\gt \pi$.

 

[Albert1]

Prove $\sqrt{2}+\sqrt{3}\gt \pi$.

Spoiler

$\begin{vmatrix}
\sqrt 3-\dfrac{\pi}{2}
\end{vmatrix}>\begin{vmatrix}
\sqrt 2-\dfrac{\pi}{2}
\end{vmatrix}=\begin{vmatrix}
\dfrac{\pi}{2}-\sqrt 2
\end{vmatrix}$
we have:
$\sqrt 3-\dfrac{\pi}{2}>\dfrac {\pi}{2}-\sqrt 2>0$
$\therefore \sqrt 3+\sqrt 2>\pi$
the proof is done

 

[anemone]

Spoiler

$\begin{vmatrix}
\sqrt 3-\dfrac{\pi}{2}
\end{vmatrix}>\begin{vmatrix}
\sqrt 2-\dfrac{\pi}{2}
\end{vmatrix}=\begin{vmatrix}
\dfrac{\pi}{2}-\sqrt 2
\end{vmatrix}$
we have:
$\sqrt 3-\dfrac{\pi}{2}>\dfrac {\pi}{2}-\sqrt 2>0$
$\therefore \sqrt 3+\sqrt 2>\pi$
the proof is done

Very well done, Albert!(Clapping)

 

[Opalg]

Spoiler

$\begin{vmatrix}
\sqrt 3-\dfrac{\pi}{2}
\end{vmatrix}>\begin{vmatrix}
\sqrt 2-\dfrac{\pi}{2}
\end{vmatrix}=\begin{vmatrix}
\dfrac{\pi}{2}-\sqrt 2
\end{vmatrix}$
we have:
$\sqrt 3-\dfrac{\pi}{2}>\dfrac {\pi}{2}-\sqrt 2>0$
$\therefore \sqrt 3+\sqrt 2>\pi$
the proof is done

[sp]I don't understand this argument. Why is $\Bigl|\sqrt 3-\dfrac{\pi}{2}\Bigr| > \Bigl|\dfrac{\pi}{2}-\sqrt 2\Bigr|$?[/sp]

 

[Albert1]

[sp]I don't understand this argument. Why is $\Bigl|\sqrt 3-\dfrac{\pi}{2}\Bigr| > \Bigl|\dfrac{\pi}{2}-\sqrt 2\Bigr|$?[/sp]

Spoiler

if $\Bigl|A\Bigr| > \Bigl|B\Bigr|$
and $\Bigl|B\Bigr| = \Bigl|C\Bigr|$
then
$\Bigl|A\Bigr| > \Bigl|C\Bigr|$

 

[anemone]

[sp]I don't understand this argument. Why is $\Bigl|\sqrt 3-\dfrac{\pi}{2}\Bigr| > \Bigl|\dfrac{\pi}{2}-\sqrt 2\Bigr|$?[/sp]

if $\Bigl|A\Bigr| > \Bigl|B\Bigr|$
and $\Bigl|B\Bigr| = \Bigl|C\Bigr|$
then
$\Bigl|A\Bigr| > \Bigl|C\Bigr|$

Spoiler

I have long thought about the validity of Albert's solution, but his argument I think is also built on the facts that he knows $\sqrt{3}>\dfrac{\pi}{2}$ and also, $\sqrt{2}<\dfrac{\pi}{2}$.

Anyway, here is my solution:

We could use the well-known inequality that says $\dfrac{22}{7}>\pi$ to assist in my method of proving, as shown below:

Note that

$7938>7921$

$2(63)^2>89^2$

$\sqrt{2}>\dfrac{89}{63}$ and

$762048>760384$

$3(504)^2>872^2$

$\sqrt{3}>\dfrac{872}{504}$,

$\therefore \sqrt{2}+\sqrt{3}>\dfrac{22}{7}>\pi$ and we're hence done.

 

[Albert1]

Spoiler

it is very easy to prove $\sqrt 3>\dfrac {\pi}{2}$ , and $\sqrt 2<\dfrac {\pi}{2}$
for :
$2<2.25=\dfrac{3^2}{4}<(\dfrac {\pi}{2})^2<\dfrac {3.2^2}{4}=2.56<3$