1.

Prove that

holds for all real numbers a, b and c

(>= means greater than or equal to)

2.

Prove that if real numbers a, b and c satisfy

then each of a, b, and c is positive

Thanks for the help guys ^_^

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- May 31st 2010, 05:28 AMJintoLynInequalities
1.

Prove that

holds for all real numbers a, b and c

(>= means greater than or equal to)

2.

Prove that if real numbers a, b and c satisfy

then each of a, b, and c is positive

Thanks for the help guys ^_^ - May 31st 2010, 08:14 AMJester
- May 31st 2010, 08:25 AMUnbeatable0
Another way for the first one: AM-GM inequality:

For the second one, consider the polynomial:

Where

Assume one of the roots, say , is negative. Put in the above equality and you'll get 0 < 0, which is absurd. - May 31st 2010, 08:32 AMBBAmp
I believe both can be solved by contradiction.

1.

Assume the opposite is true:

for all integers.

or for simplicity

we can instantly see by inspection that any integer does not work since is false.

2.

From and assuming that inequality is correct you can instantly get .

If you start off by assuming one or more variable is less than zero and also assume this set of equations work for all positive or negative integers (depending upon which variable you assume is less than or greater than zero) you will run into contradictions

EDIT: I didn't see the first two replies prior to posting this.. you guys are too quick for me :) - May 31st 2010, 08:34 AMJester
Here's the second. From the third equation so . Also due to the sign of the third equation either all are positive or two of the three are negative. We will show only the first case is true. Assume that and are negative. Let and so and .

Thus

and

.

Multiplying the inequalities gives

which leads to a contradiction since .

Too slow but with details (Rofl) - May 31st 2010, 03:42 PMArchie Meade
- May 31st 2010, 04:10 PMArchie Meade