Minimum of x+1/x (Perimeter Square < Perim equal area rects)

[CynicusRex]

Homework Statement

Gelfand - Algebra p.115 problem 264:
Prove that a square has the minimum perimeter of all rectangles having the same area.
Hint. Use the result of the preceding problem.

Homework Equations

Preceding problem: Prove that a square has the maximum area of all rectangles having the same perimeter.

Perimeter of a square = Perimeter of a rectangle with an equal perimeter: $${4x} = 2(x-r)+2(x+r) = 2x-2r+2x+2r = 4x$$
Area of that square: x²
Area of that rectangle: (x-r)(x+r) = x²+rx-rx-r² = x²-r²

Since r² is always positive, then x² > x²-r².

The Attempt at a Solution

My initial attempt, I did coincidentally try with the result of the preceding problem. Let's take the area of a square and let it be equal to the area of a rectangle: x² = x² - r²
The perimeter of that square: 4x
The perimeter of that rectangle: 2(x-r)+2(x+r) = 2x-2r+2x+2r = 4x
So if the area of a square and a rectangle are equal, the perimeter of that rectangle is equal to the perimeter of the square. Then it dawned on me that x² = x² - r² is impossible.

Second try:
The area of a square = all rectangles with the same area: $$x² = (ax)(\frac{1}{a}x)$$
The perimeter of that square: 4x
The perimeter of those rectangles: $$2ax+\frac{2}{a}x$$
$$2x(a+\frac{1}{a})$$

Now I got to prove, for a positive a not equal to 1, that: $$a+\frac{1}{a} > 2 $$

When that's proven, it shows that a square always has the minimum perimeter of all rectangles having the same area. (when a=1, it is the same square, not a rectangle)

Here my first try was a proof by contradiction.
$$a+\frac{1}{a} < 2$$
$$\frac{1}{a} < 2 - a$$
$$1 < 2a - a²$$
$$0 < 2a - a² - 1$$
Multiplied both sides by -1
$$0 > a² - 2a + 1$$
$$0 > (a-1)²$$
Which is false because the square is always positive. So: $$4x < 2x(a+\frac{1}{a})$$
Could someone lead me in the right direction towards finding other, perhaps easier proofs?

 

[Ray Vickson]

Homework Statement

Gelfand - Algebra p.115 problem 264:
Prove that a square has the minimum perimeter of all rectangles having the same area.
Hint. Use the result of the preceding problem.

Homework Equations

Preceding problem: Prove that a square has the maximum area of all rectangles having the same perimeter.

Perimeter of a square = Perimeter of a rectangle with an equal perimeter: $${4x} = 2(x-r)+2(x+r) = 2x-2r+2x+2r = 4x$$
Area of that square: x²
Area of that rectangle: (x-r)(x+r) = x²+rx-rx-r² = x²-r²

Since r² is always positive, then x² > x²-r².

The Attempt at a Solution

My initial attempt, I did coincidentally try with the result of the preceding problem. Let's take the area of a square and let it be equal to the area of a rectangle: x² = x² - r²
The perimeter of that square: 4x
The perimeter of that rectangle: 2(x-r)+2(x+r) = 2x-2r+2x+2r = 4x
So if the area of a square and a rectangle are equal, the perimeter of that rectangle is equal to the perimeter of the square. Then it dawned on me that x² = x² - r² is impossible.

Second try:
The area of a square = all rectangles with the same area: $$x² = (ax)(\frac{1}{a}x)$$
The perimeter of that square: 4x
The perimeter of those rectangles: $$2ax+\frac{2}{a}x$$
$$2x(a+\frac{1}{a})$$

Now I got to prove, for a positive a not equal to 1, that: $$a+\frac{1}{a} > 2 $$

When that's proven, it shows that a square always has the minimum perimeter of all rectangles having the same area. (when a=1, it is the same square, not a rectangle)

Here my first try was a proof by contradiction.
$$a+\frac{1}{a} < 2$$
$$\frac{1}{a} < 2 - a$$
$$1 < 2a - a²$$
$$0 < 2a - a² - 1$$
Multiplied both sides by -1
$$0 > a² - 2a + 1$$
$$0 > (a-1)²$$
Which is false because the square is always positive. So: $$4x < 2x(a+\frac{1}{a})$$
Could someone lead me in the right direction towards finding other, perhaps easier proofs?

Use the arithmetic-geometric mean inequality. When are the two types of means equal?