What typo?

As for you avatar, what exactly is it?

well... the intricate design was created by joining lots of circles together in a chain, then deleting the alternate segments and filling in the middle, as for what it IS, i have no idea, but i think it looks quite cool :X

]]>As for you avatar, what exactly is it?

]]>woow brilliant ! ... just one thing :

i meant the - how did it become + ?!

following what else he wrote, it seems to be a typo

and off topic? who likes my avatar, i think it looks rather cool

]]>xy < \frac{x^2 - y^2}{2}

how did it become :

\frac{xy}{2} < \frac{x^2 + y^2}{4}

i meant the - how did it become + ?!

sorry really sorry i bothered u alot !

]]>Since x > y, it must be that 0 < x - y and 0 < (x - y)(x - y). So...

So

Edit: Ok, I checked it over like 500 times. That *has* to be right.

Since x and y are both positive, it must be that:

-xy < x² + y²

So:

-xy + 2xy < x² + 2xy + y²

And:

xy < x² + 2xy + y² = s²

]]>( XY + 4 (( X + y ) / 2 )) / 2 = ( X + y ) ² / 4 ... how is that ?

( X + Y ) ² = X² + 2 XY + Y ²]]>

I like your "read from the middle outwards" comparison.

Another approach would be to compare the areas for a square and for a rectangle that is just "Δx" from being a square.

]]>Area of square: s²

Area of rectangle: xy

What we wish to conclude is xy < s².

Thus, xy < s²

]]>is there any other way to prove it ?]]>

The function is: A = Pw/2 - w²

It's derivative is: A' = P/2 - 2w

(I am not sure if you know about derivatives, but it is part of calculus.)

Now, for the sake of making a graph I set P=10, and plotted the functions together: Plot of 10*x/2-x^2 vs 10/2-2*x (x means w)

Notice that the "10/2 - 2w" line shows you the slope of the "10*w/2 - w²" line?

Also notice that where the "10/2 - 2w" crosses the zero line is exactly where the "10*w/2 - w²" changes direction?

And, as it turns out, that is the maximum value for the "A = 10*w/2 - w²" function.

That is because a curve flattens out (slope=0) on the top (or bottom) when it reaches a maximum (or minimum) value.

So I used this idea to find the largest value of A. That is, A will be at its peak when A'=0.

There can be pitfalls with this - you don't actually know if it is a maximum or minimum, or the curve may later come back and go even further, or it may stay flattened out, etc. But with a simple function like this it works nicely.

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