Problem 36

Algebra Level 5

( a b + b a + 1 ) ( 1 a − 1 b ) 2 a 2 b 2 + b 2 a 2 − ( a b + b a ) \large \dfrac{\left(\frac{a}{b}+\frac{b}{a}+1\right)\left(\frac{1}{a}-\frac{1}{b}\right)^2}{\frac{a^2}{b^2}+\frac{b^2}{a^2}-\left(\frac{a}{b}+\frac{b}{a}\right)}

Given that a a and b b are distinct positive reals satisfying 4 a + b + a b = 1 4a+b+\sqrt{ab}=1 .

Find the minimum value of the expression above.

See my set of inequality problems .


The answer is 25.

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Son Nguyen by Son Nguyen , 20, Vietnam

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2 solutions

P C
Mar 11, 2016

Call the expression A A , set t = a b + b a t=\frac{a}{b}+\frac{b}{a} and we get A = ( t + 1 ) ( a − b a b ) 2 t 2 − t − 2 = ( a − b a b ) 2 t − 2 A=\frac{(t+1)\big(\frac{a-b}{ab}\big)^2}{t^2-t-2}=\frac{\big(\frac{a-b}{ab}\big)^2}{t-2} Now we see that t − 2 = a b + b a − 2 = ( a − b ) 2 a b t-2=\frac{a}{b}+\frac{b}{a}-2=\frac{(a-b)^2}{ab} , so A = a b ( a − b ) 2 [ a b ( a − b ) ] 2 = 1 a b A=\frac{ab(a-b)^2}{[ab(a-b)]^2}=\frac{1}{ab} Applying the arithmetic mean geometric mean inequality , we get 4 a + b ≥ 4 a b 4a+b\geq 4\sqrt{ab} ∴ 1 ≥ 5 a b \therefore 1\geq 5\sqrt{ab} ⇔ A = 1 a b ≥ 25 \Leftrightarrow A=\frac{1}{ab}\geq 25 The equality holds when 4 a = b 4a=b , or a = 1 10 , b = 2 5 a = \dfrac1{10}, b= \dfrac2{5} .

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Exactly Same Way.

Kushagra Sahni - 5 years, 3 months ago
Aakash Khandelwal
Mar 12, 2016

To get the answer following procedures must be followed. 1. 1. Rearrange the following expression to get the value of expression as 1 / a b 1/ab . 2. 2. Apply A M ≥ G M o n 4 a AM\geq GM on 4a and b b . That is ( 4 a + b ) ≥ 4 × ( a b ) (4a+b)\geq 4\times \sqrt(ab) . Hence maximum value of a b ab is 1 / 25 1/25 . 3. 3. Therefore minimum value of given expression is
25 \boxed{25}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Now this is the solution to get answer. But while solving the problem I got stuck somewhere checkout? Now if we apply AM-GM as 4 a + b + ( a b ) ≥ 3 × ( 4 × ( a b ) 3 / 2 ) 1 / 3 4a + b + \sqrt(ab)\geq 3\times (4\times (ab)^{3/2})^{1/3} We get minimum value of expression as 22.6785 \boxed {22.6785} . Where am I wrong?

Aakash Khandelwal - 5 years, 3 months ago

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If you use AM-GM for 4 a ; b ; a b 4a;b;\sqrt{ab} , you'll get the equality case when 4 a = b = a b 4a=b=\sqrt{ab} which can't be happened

P C - 5 years, 3 months ago

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Got it. You removed my doubt.

Aakash Khandelwal - 5 years, 3 months ago

Ah yes, the classic "But equality cannot occur".

Always remember that you've only found a Lower Bound . To show that it is the minimum, you must show that equality can exist, so that it's the greatest lower bound.

Calvin Lin Staff - 5 years, 3 months ago

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