Question

Explain why
`√(xy) \leq (x+y)/(2)` for any two positive real numbers x and y,
`x,y \\eq 0` (The geometric mean is always less than the arithmetic mean.)

Answer 1

`\large\rm (x+y)/(2)\ge √(xy)`

Squaring each side gives us,

`\large\rm ((x+y)^2)/(4)\ge xy`

Multiply by 4,

`\large\rm (x+y)^2\ge 4xy`

and subtract,

`\large\rm (x+y)^2-4xy\ge 0`


Expand out the square,

`\large\rm x^2+2xy+y^2-4xy\ge0`

Combine like-terms,

`\large\rm x^2-2xy+y^2\ge0`

This can now be written as a different perfect square,

`\large\rm (x-y)^2\ge0`

Recall that a squared value is always positive. So (x-y)^2 is always greater than zero.

That should "explain it" sufficiently unless you need to do some sort of formal proof.