Inequality questionfest 2

Algebra Level 4

Let a a , b b , c c , and d d be positive reals such that s = a + b + c + d = 1 s=a+b+c+d=1 . Find the minimum value of

( s − a ) ( s − b ) ( s − c ) ( s − d ) a b c d . \large \dfrac{(s-a)(s-b)(s-c)(s-d)}{abcd} .


The answer is 81.

Madhav Srirangan by madhav srirangan , 20, India

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4 solutions

X = ( s − a ) ( s − b ) ( s − c ) ( s − d ) a b c d = ( 1 − a ) ( 1 − b ) ( 1 − c ) ( 1 − d ) a b c d = 1 − ( a + b + c + d ) + ( a b + a c + a d + b c + b d + c d ) − ( a b c + a b d + a c d + b c d ) + a b c d a b c d = 1 − 1 + ( a b + a c + a d + b c + b d + c d ) − ( a b c + a b d + a c d + b c d ) + a b c d a b c d = ( a b + a c + a d + b c + b d + c d ) − ( a b c + a b d + a c d + b c d ) + a b c d a b c d By AM-GM inequality ≥ 6 ( a b c d ) 1 2 − 4 ( a b c d ) 3 4 + a b c d a b c d Equality occurs when a = b = c = d = 1 4 ≥ 6 ( 1 4 4 ) − 1 2 − 4 ( 1 4 4 ) − 1 4 + 1 ≥ 6 ( 16 ) − 4 ( 4 ) + 1 ≥ 81 \begin{aligned} \mathscr X & = \frac {(s-a)(s-b)(s-c)(s-d)}{abcd} \\ & = \frac {(1-a)(1-b)(1-c)(1-d)}{abcd} \\ & = \frac {1-(a+b+c+d)+(ab + ac + ad + bc + bd + cd) - (abc + abd + acd + bcd) + abcd}{abcd} \\ & = \frac {1-1+(ab + ac + ad + bc + bd + cd) - (abc + abd + acd + bcd)+ abcd}{abcd} \\ & = \frac {(\color{#3D99F6}{ab + ac + ad + bc + bd + cd}) - (\color{#3D99F6}{abc + abd + acd + bcd})+ abcd}{abcd} \quad \quad \small \color{#3D99F6}{\text{By AM-GM inequality}} \\ & \ge \frac {\color{#3D99F6}{6(abcd)^\frac 12} - \color{#3D99F6}{4(abcd)^\frac 34}+ abcd}{abcd} \quad \quad \small \color{#3D99F6}{\text{Equality occurs when }a=b=c=d=\frac 14} \\ & \ge 6 \left(\color{#3D99F6}{\frac 1{4^4}} \right)^{-\frac 12} - 4\left(\color{#3D99F6}{\frac 1{4^4}} \right)^{-\frac 14}+ 1 \\ & \ge 6 \left(16 \right) - 4\left( 4 \right) + 1 \\ & \ge \boxed{81} \end{aligned}

4 Helpful 1 Interesting 2 Brilliant 0 Confused

Sir, I read a little about AM-GM and I understand a little bit of it, however I have seen that in most of these types of questions, the minimum occurs when all variables are equal. Is this a coincidence or just a result of this inequality? I am asking this because I am preparing for a MCQ exam, so I may not have time to completely solve it, and if this is a correct way, it may help me solve these type of questions quickly. Thanks in advance!

Vinayak Srivastava - 10 months, 2 weeks ago

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I am no expert in this. I also made quite a number of mistakes there and then. Best that you read up for it and try more problems in it. Usually three valuables (a), b b , and c c are used. They must be independent for them to work. And when you have establish an inequality using the theorem, an actual solution must exist before the solution is valid. As shown in the solution above. "By AM-GM inequality" establishes the inequality, but "Equality must occur" with actual values of a a , b b , and c c , before the inequality is valid.

Chew-Seong Cheong - 10 months, 2 weeks ago

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Thanks for your advice! I actually solved it directly like @Novril Razenda has shown. I think I will read it once or twice before the exam, so that I remember what happens when. Thank you!

Vinayak Srivastava - 10 months, 2 weeks ago
Hải Trung Lê
Aug 5, 2016

( s − a ) ( s − b ) ( s − c ) ( s − d ) a b c d = ( b + c + d ) ( a + c + d ) ( a + b + d ) ( a + b + c ) a b c d ≥ ( 3 b c d 3 ) ( 3 a c d 3 ) ( 3 a b d 3 ) ( 3 a b c 3 ) a b c d (AM-GM inequality) = 81 a b c d a b c d = 81 Equality occurs when a = b = c = d = 1 4 \begin{aligned} & \ \quad \frac{\color{#D61F06}{(s-a)}\color{#3D99F6}{(s-b)}\color{#20A900}{(s-c)}\color{magenta}{(s-d)}}{\color{#69047E}{abcd}}\\ &= \frac{\color{#D61F06}{(b+c+d)}\color{#3D99F6}{(a+c+d)}\color{#20A900}{(a+b+d)}\color{magenta}{(a+b+c)}}{\color{#69047E}{abcd}}\\ &\geq \frac{\color{#D61F06}{(3\sqrt[3]{bcd})}\color{#3D99F6}{(3\sqrt[3]{acd})}\color{#20A900}{(3\sqrt[3]{abd})}\color{magenta}{(3\sqrt[3]{abc})}}{\color{#69047E}{abcd}} \quad \quad \quad \text{(AM-GM inequality)}\\ &= \frac{81\color{#69047E}{abcd}}{\color{#69047E}{abcd}}\\ &= \boxed{81} \end{aligned}\\ \text{Equality occurs when }a=b=c=d=\frac{1}{4}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Mohammed Imran
Apr 3, 2020

Let f ( x ) = 1 − x x f(x)=\frac{1-x}{x} . Since f ( x ) f(x) is a convex function, we have f ( a + b + c + d 4 ) ≤ f ( a ) + f ( b ) + f ( c ) + f ( d ) 4 f(\frac{a+b+c+d}{4}) \leq \frac{f(a)+f(b)+f(c)+f(d)}{4} so, we have f ( a ) + f ( b ) + f ( c ) + f ( d ) ≥ 12 f(a)+f(b)+f(c)+f(d) \geq 12 but we want maximum value for f ( a ) × f ( b ) × f ( c ) × f ( d ) f(a) \times f(b) \times f(c) \times f(d) . So, applying A.M.-G.M. inequality on f ( a ) , f ( b ) , f ( c ) , f ( d ) f(a), f(b), f(c), f(d) , we have f ( a ) + f ( b ) + f ( c ) + f ( d ) ≥ 4 ( f ( a ) × f ( b ) × f ( c ) × f ( d ) ) ( 1 4 ) ≥ 12 f(a)+f(b)+f(c)+f(d) \geq 4(f(a) \times f(b) \times f(c) \times f(d))^(\frac{1}{4}) \geq 12 so, the minimum value is 3 4 = 81 3^4=\boxed{81}

0 Helpful 0 Interesting 1 Brilliant 0 Confused
Novril Razenda
Jul 15, 2016

use AM-GM we can find ( s − a ) ( s − b ) ( s − c ) ( s − d ) a b c d = 3 4 4 4 1 4 4 4 = 3 4 = 81 \large \frac{(s-a)(s-b)(s-c)(s-d)}{abcd}=\frac{\frac{3^4}{4^4}}{\frac{1^4}{4^4}}=3^4=81

1 Helpful 0 Interesting 0 Brilliant 0 Confused

sir please elaborate

Deepansh Jindal - 4 years, 10 months ago

Yes please elaborate

Mohammed Imran - 1 year, 2 months ago

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