# some inequalities

• Mar 25th 2007, 12:41 PM
sacwchiri
some inequalities
started college and havent been doing so good and well i have this test soon and im having trouble with some questions... (i appologize in advance i have to translate from spanish... so if something sounds funny well ask me)

1) Determine the dimensions of the rectangle with the greatest area which perimeter is 8.

2) If x, y > 0. prove that:

2/((1/x)+(1/y)) <= sqrt(xy) <= (x+y)/2

3) If a, b, c are positive real numbers and different of each other, demonstrate that:

(a + b + c) / 3 > ((ab + bc + ca)/3)^(1/2) > (abc)^(1/3)
• Mar 25th 2007, 12:59 PM
Jhevon
Quote:

Originally Posted by sacwchiri
started college and havent been doing so good and well i have this test soon and im having trouble with some questions... (i appologize in advance i have to translate from spanish... so if something sounds funny well ask me)

1) Determine the dimensions of the rectangle with the greatest area which perimeter is 8.

This is an optimization problem. See the diagram below

The Calculus way to do this problem.

we have P = 2x + 2y = 8 ............P is the perimeter
and we want the area A = xy to be a maximum

from P, we have
2x + 2y = 8
=> x = 4 - y ..........solved for x

=> A = xy = (4 - y)y = 4y - y^2

for max area, we want A' = 0
=> A' = 4 - 2y = 0
=> 4 - 2y = 0
=> y = 2

but x = 4 - y
=> x = 4 - 2
=> x = 2

so for our rectangle, x = y = 2, so we want a 2 x 2 rectangle

The Precalculus way to do this problem

we have P = 2x + 2y = 8 ............P is the perimeter
and we want the area A = xy to be a maximum

from P, we have
2x + 2y = 8
=> x = 4 - y ..........solved for x

=> A = xy = (4 - y)y = 4y - y^2

For max A, we want the vertex of the parabola 4y - y^2 = 0

vertex occurs when y = -b/2a
=> we want y = -4/2(-1) = -4/-2 = 2
so y = 2

but x = 4 - y
=> x = 4 - 2
=> x = 2

so for our rectangle, x = y = 2, so we want a 2 x 2 rectangle

got it?
• Mar 25th 2007, 01:07 PM
ThePerfectHacker
Quote:

Originally Posted by sacwchiri
1) Determine the dimensions of the rectangle with the greatest area which perimeter is 8.

That rectangle is actually a square.
Quote:

2) If x, y > 0. prove that:

2/((1/x)+(1/y)) <= sqrt(xy) <= (x+y)/2
There is nothing to prove.
This is the
HM-GM-AM
Inequality.
Quote:

3) If a, b, c are positive real numbers and different of each other, demonstrate that:

(a + b + c) / 3 > ((ab + bc + ca)/3)^(1/2) > (abc)^(1/3)
By Cauchy-Swartz Inequality (my personal favorite)
When a,b,c are different.

a^2+b^2+c^2 > ab+bc+ac

a^2+b^2+c^2+2ab+2ac+2bc > 3ab+3bc+3ac

(a^2+2ab+b^2)+2c(a+b)+c^2 > 3ab+3bc+3ac

(a+b)^2+2(a+b)c+c^2 > 3ab+3bc+3ac

(a+b+c)^2 > 3ab+3bc+3ac

(a+b+c)^2/9 > (ab+bc+ac)/3

Take square roots,

(a+b+c)/3 > ((ab+bc+ac)/3)^(1/2)
Q.E.D.
• Mar 25th 2007, 01:17 PM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
This is the
HM-GM-AM
Inequality.

what's that? where can i find a proof for it?
• Mar 25th 2007, 01:20 PM
sacwchiri
axioms and reals
i got what Jhevon said but for theperfecthcaker what is the hm-gm-am inequality... i should have add im not any good at math...

and maybe while you are at it explain the Cauchy-Swartz Inequality... cos i dont get that part...

thanx :)
• Mar 25th 2007, 01:32 PM
ThePerfectHacker
Heir
• Mar 25th 2007, 01:41 PM
sacwchiri
sooo what you mean is that
a^2+b^2+c^2 will always be >= ab+bc+ac???

and then i get kinda lost...
(sorry for anoying but this is really helping me out now :o)
• Mar 25th 2007, 01:47 PM
ThePerfectHacker
Quote:

Originally Posted by sacwchiri
sooo what you mean is that
a^2+b^2+c^2 will always be >= ab+bc+ac???

and then i get kinda lost...
(sorry for anoying but this is really helping me out now :o)

a^2+b^2+c^2 > = ab+bc+ac when a,b,c>0

Why?

Explanation.
-------------

You can write,
(a^2+b^2+c^2)(b^2+c^2+a^2) > = (ab+bc+ac)^2
By Cauchy-Swartzh Inequality.
Thus,
(a^2+b^2+c^2)^2 >= (ab+bc+ac)^2
Take square roots (and note that a,b,c>0)

a^2+b^2+c^2 >= ab+bc+ac
Since,
a != b!=c
We cannot have equality and hence,
a^2+b^2+c^2 > ab+bc+ac