If $x + y + z \leq 3$ then $1 / x + 1 / y + 1 / z \geq 3$

By Alex Holyoake on Monday, February 18, 2002 - 09:21 am:

If there are 3 positive numbers,

x

y

z

such that

x + y + z \leq 3

, how can it be proved that

1 / x + 1 / y + 1 / z \geq 3

?
Help!

By Alex Holyoake on Monday, February 18, 2002 - 01:14 pm:

x is not necessarily equal to y which is not necessarily equal to z, x, y and z are all greater than 0. so for example you could have x = 2.8 y = 0.1 and z = 0.1

By David Loeffler on Monday, February 18, 2002 - 01:26 pm:

Ah! The simplest proof I know uses a result called the Cauchy-Schwarz inequality. This says that

$(x_{1}^{2} + x_{2}^{2} + x_{3}^{2}) (y_{1}^{2} + y_{2}^{2} + y_{3}^{2}) \geq (x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3})^{2}$

for any real numbers $x_{1 \dots 3}$ and $y_{1 \dots 3}$ .

In your case just put $x_{1} = \sqrt{x}$ , $y_{1} = 1 / \sqrt{x}$ , and similarly $x_{2} = \sqrt{y}$ , $y_{2} = 1 / \sqrt{y}$ etc. Then we get

$(x + y + z) (1 / x + 1 / y + 1 / z) \geq (1 + 1 + 1)^{2} = 9$ .

So if $x + y + z \leq 3$ , $1 / x + 1 / y + 1 / z \geq 3$ , and so on.

David

By Alex Holyoake on Monday, February 18, 2002 - 01:41 pm:

Thanks everso!
Just two questions, how do you deduce the last line of your proof from the second to last line, and also what is the proof behind the cauchy schwartz equation

By David Loeffler on Monday, February 18, 2002 - 03:44 pm:

Well, we have shown that two numbers

(x + y + z)

and

(1 / x + 1 / y + 1 / z)

have product at least 9; so they can't both be less than 3.

To prove the inequality is a little difficult; there are various proofs. The standard one is like this:

Consider the quadratic

$(x_{1} + t y_{1})^{2} + (x_{2} + t y^{2})^{2} + (x_{3} + t y^{3})^{2}$

If you expand it out, it becomes

$t^{2} (x_{1}^{2} + x_{2}^{2} + x_{3}^{2}) + 2 t (x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3}) + (y_{1}^{2} + y_{2}^{2} + y_{3}^{2})$

Now, we know this can't possibly be negative, because it's a sum of three squares and squares are always positive.

If we have a quadratic of the form $a t^{2} + b t + c$ , then this is zero when $t = (- b \pm \sqrt{b^{2} - 4 a c}) / 2 a$ . You may have come across this at school; it's a standard formula.

Right: here is the (slightly) clever bit. If the quadratic in $t$ above has two distinct roots, it must be negative in between them. We know this can't happen because it's a sum of squares. So the formula must give either only one root, which can only happen if $b^{2} - 4 a c = 0$ , or no real roots at all, which happens if $b^{2} - 4 a c < 0$ .

So, let's calculate this quantity $b^{2} - 4 a c$ (it's called the discriminant) for the quadratic above. It turns out to be

$4 (x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3})^{2} - 4 (x_{1}^{2} + x_{2}^{2} + x_{3}^{2}) (y_{1}^{2} + y_{2}^{2} + y_{3}^{2})$

So this expression must be $\leq 0$ , i.e.

$(x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3})^{2} \leq (x_{1}^{2} + x_{2}^{2} + x_{3}^{2}) (y_{1}^{2} + y_{2}^{2} + y_{3}^{2})$

which is the C-S inequality. David

By Michael Doré on Monday, February 18, 2002 - 04:09 pm:

Alternatively you can use the AM-GM inequality. AM-GM says that for any positive numbers, the arithmetic mean is greater than or equal to their geometric mean. So:

(x + y + z)/3 > = (xyz)^1/3

(1/x + 1/y + 1/z)/3 > = (xyz)^-1/3

Multiplying these together we get:

(x + y + z)(1/x + 1/y + 1/z)/9 > = 1

and then the solution follows as in David's proof.