Minimum 3

Algebra Level 4

If $a$ and $b$ are positive numbers satisfying $ab (a+b+1)=25$ , find the minimum value of $(a+b)(b+1)$ .

The answer is 10.

3 solutions

Jun Shin
Mar 2, 2016

$(ab)$ * $(a+b+1)$ =25 dividing both sides by $a$ ,
$b$ * $(a+b+1)$ = $\frac{25}{a}$ .
$(a+b)$ * $(b+1)$ = $a$ + $b$ * $(a+b+1)$ = $a$ + $\frac{25}{a}$ .
By the AM-GM inequality, $a$ + $\frac{25}{a}$ ≥ 2√ $(a*(\frac{25}{a}$ ))= $10$ .
The minimum value is $10$

For completeness of solution you should also mention the equality case. In this case it occurs when $a=5, b=-3+\sqrt{14}$ .

Rishabh Jain - 5 years, 3 months ago

That's a good point! We don't know if $b$ is positive as yet :)

Calvin Lin Staff - 5 years, 3 months ago

Great problem!

Note: To start a new line, just leave 3 empty spaces at the end of your line and hit enter. I've edited your solution to separate out your statements. You can click "edit" to see the changes that I made.

Calvin Lin Staff - 5 years, 3 months ago

Thanks for helping me edit!

Jun Shin - 5 years, 3 months ago

Prakhar Bindal
Mar 3, 2016

Firstly Apply AM-GM On ab(a+b+1 ) by taking them as three different terms.

Then Apply AM-GM On ab,a/3,a/3,a/3,b^2 and b to get the desired result and combine the two resulting inequalities to get required answer as 10.

hi, please explain what you did in the second step. i couldn't understand what you did there. thanks.

Lipsa Kar - 5 years, 3 months ago

i applied AM GM On those 6 quantities and my motivation to take these was to make the powers of a and b equal in the Geometric mean side in both the inequalities so i can compare them easily

Prakhar Bindal - 5 years, 3 months ago

also, can you please see my method? I'm getting a different answer here.what i did is (a+b)(b+1)=ab+a+b^2+b=ab+b^2+25/ab -1=[(ab-5)^2]/ab+10+b^2-1. to get minimum, ab=5, solving, i got b=2-i and a=2+i (complex numbers). so minimum is 12-4i. modulus is 4x10^(1/2). so, the answer is 12.64

Lipsa Kar - 5 years, 3 months ago

Firstly as i see a and b are positive real numbers as specified in the question so no need to go into complex numbers buisness. for your second point you cannot maximise like this as b is also a variable not a constant so b can vary also.there might be some other case for which maxima is attained

Prakhar Bindal - 5 years, 3 months ago

yes, the question asks for +ve real but when i went to solve it, i got complex, which varied with what the question stated. also, why do you say the minimum cant be found out in that method? b can be assumed as a constant here, there are no restrictions.

Lipsa Kar - 5 years, 3 months ago

@Lipsa Kar – I Mean to say just making perfect square zero does not ensure minima to be attained as far as b can vary. for example

(ab-5)^2 + b^2 . if you want to minimse this we may say that ab= 5 but it is not always true.

say i take ab = 5.0000000000001 and correspondingly we might get a value of b which makes the overall expression lesser than what is obtained when ab= 5.

Prakhar Bindal - 5 years, 3 months ago

@Prakhar Bindal – hmm, my method doesnt give the minima but i dont know why. plus, both the terms above are squares and hence, positive terms. taking ab=5.000001 should also yield a quantity to be added, right?

Lipsa Kar - 5 years, 3 months ago

@Lipsa Kar – Yeah you are right but that quantity may be less than what is added when ab = 5 right?

Prakhar Bindal - 5 years, 3 months ago

Can you be more explicit in your solution? If you list out the eequation directly, others can see why your solution works.

Calvin Lin Staff - 5 years, 3 months ago

I Gede Arya Raditya Parameswara
Mar 3, 2016

(ab)*(a+b+1)=25-1 a=4 b=1 (4+1)(1+1)=5×2=10 so the minimum value of (a+b)(b+1)=10

Why must that be the case?

Calvin Lin Staff - 5 years, 3 months ago

Minimum 3

The answer is 10.

3 solutions

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