I am working on a howework question, trying to prove the following: 5 a + b >

Adelyn Rodriguez

Adelyn Rodriguez

Answered question

2022-05-11

I am working on a howework question, trying to prove the following:
5 a + b > 4 a b , ,
where a and b are positive real numbers.
I've tried multiplying expression by a b , squaring both sides of the equation so far. In both cases, after re-factoring, I could not conclude that inequality holds. Can someone point me in the right direction?

Answer & Explanation

percolarse2rzd

percolarse2rzd

Beginner2022-05-12Added 17 answers

Step 1
In general you have to be a bit careful about squaring inequalities, but here there’s no problem: since we’re assuming that a and b are positive, 5 a + b is positive, and therefore 5 a + b > 4 a b if and only if ( 5 a + b ) 2 > 16 a b , i.e., if and only if 25 a 2 6 a b + b 2 > 0 . Now think of b 2 6 a b + 25 a 2 as a quadratic in b and complete the square: b 2 6 a b + 25 a 2 = ( b 3 a ) 2 + 16 a 2 . Clearly ( b 3 a ) 2 0 , and since a is positive, a 2 > 0 , so ( b 3 ) 2 + 16 a 2 > 0 .
Retracing our steps, we see that ( 5 a + b ) 2 > 16 a b and hence that 5 a + b = ( 5 a + b ) 2 > 4 a b .
enclinesbnnbk

enclinesbnnbk

Beginner2022-05-13Added 5 answers

Step 1
For any real x and y you have
2 x y = x 2 + y 2 ( x y ) 2
taking x = 2 a and y = b you get
2 ( 2 a b ) = 4 a + b ( 2 a b ) 2 4 a + b < 5 a + b

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?