Both maxima and minima

Algebra Level 4

Non-negative real numbers a a , b b , c c , and d d are such that a + b + c + d = 100 a + b + c + d = 100 . Find the maximum value of the sum below.

a b + 7 3 + b c + 7 3 + c d + 7 3 + d a + 7 3 \sqrt [ 3 ]{ \frac { a }{ b + 7 } } + \sqrt [ 3 ]{ \frac { b }{ c + 7 } } + \sqrt [ 3 ]{ \frac { c }{ d + 7 } } + \sqrt [ 3 ]{ \frac { d }{ a + 7 } }

8 13 3 \frac { 8 }{ \sqrt [ 3 ]{13 } } 10 7 3 \frac { 10 }{ \sqrt [ 3 ]{ 7 } } 7 8 3 \frac { 7 }{ \sqrt [ 3 ]{ 8 } } 8 7 3 \frac { 8 }{ \sqrt [ 3 ]{ 7 } } . 7 13 3 \frac { 7 }{ \sqrt [ 3 ]{ 13 } }

Priyanshu Mishra by Priyanshu Mishra , 20, India

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1 solution

Priyanshu Mishra
Jul 29, 2019

By using Hölder's inequality:

S = ( ∑ c y c a 6 ⋅ a 6 b + 7 3 ) 3 ≤ ∑ c y c ( a 6 ) 3 ⋅ ∑ c y c ( a 6 ) 3 ⋅ ∑ c y c ( 1 b + 7 3 ) 3 = ( ∑ c y c a ) 2 ∑ 1 b + 7 . \large\ { S = \left(\displaystyle \sum_{cyc} { \frac { \sqrt [ 6 ]{ a } \cdot \sqrt [ 6 ]{ a } }{ \sqrt [ 3 ]{ b + 7 } } } \right) }^{ 3 } \le \displaystyle \sum_{cyc} { { \left( \sqrt [ 6 ]{ a } \right) }^{ 3 } \cdot \displaystyle \sum_{cyc} { { \left( \sqrt [ 6 ]{ a } \right) }^{ 3 } } \cdot \displaystyle \sum_{cyc} { { \left( \frac { 1 }{ \sqrt [ 3 ]{ b + 7 } } \right) }^{ 3 } } } = { \left( \displaystyle \sum_{cyc} { \sqrt { a } } \right) }^{ 2 }\sum { \frac { 1 }{ b + 7 } } .

Notice that :

( x − 1 ) 2 ( x − 7 ) 2 x 2 + 7 ≥ 0 ⟺ x 2 − 16 x + 71 ≥ 448 x 2 + 7 \large\ \frac { { \left( x - 1 \right) }^{ 2 }{ \left( x - 7 \right) }^{ 2 } }{ { x }^{ 2 } + 7 } \ge 0 \Longleftrightarrow { x }^{ 2 } - 16x + 71\ge \frac { 448 }{ { x }^{ 2 } + 7 }

yields

∑ 1 b + 7 ≤ 1 448 ∑ ( b − 16 b + 71 ) = 1 448 ( 384 − 16 ∑ b ) = 48 − 2 ∑ b 56 . \large\ \displaystyle \sum { \frac { 1 }{ b + 7 } \le \frac { 1 }{ 448 } \sum { \left( b - 16\sqrt { b } + 71 \right) } = \frac { 1 }{ 448 } \left( 384 - 16\sum { \sqrt { b } } \right) } = \frac { 48 - 2\sum { \sqrt { b } } }{ 56 }.

Finally,

S 3 ≤ 1 56 ( ∑ a ) 2 ( 48 − ∑ 2 a ) ≤ 1 56 ( ∑ a + ∑ a + ( 48 − ∑ 2 a ) 3 ) 3 = 512 7 \large\ { S }^{ 3 } \le \frac { 1 }{ 56 } { \left( \displaystyle \sum { \sqrt { a } } \right) }^{ 2 }\left( 48 - \sum { 2\sqrt { a } } \right) \le \frac { 1 }{ 56 } { \left( \frac {\displaystyle \sum { \sqrt { a } + \displaystyle \sum { \sqrt { a } + \left( 48 - \displaystyle \sum { 2\sqrt { a } } \right) } } }{ 3 } \right) }^{ 3 } = \boxed {\frac { 512 }{ 7 }}

by the AM-GM inequality.

This bound is achieved when ( a , b , c , d ) (a, b, c, d) is a cyclic permutation of ( 1 , 49 , 1 , 49. ) (1, 49, 1, 49.)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

These inequalities are giving us an upper bound. You haven't shown that the bound can be achieved.
Three of the bounds offered as solutions are clearly too low, as demonstrated by a=b=c=d=25. But even if I let a,c-> 0 and b = 50-a, d=50-c, I get nowhere near this number. I find it harder to get higher than ~3.85187. Generally speaking, it seems that we'd want most of the weight on two non-consecutive terms. Other options are giving lower sums (work not shown.)

Richard Desper - 1 year, 10 months ago

how did you get that idea of choosing (x-1)^2(x-7)^2/(x^2+7). you want something like 1/(b+7)<= some quadratic function of squareroot(b). this is adhoc. only thing i can see is that you wanted to get rid of squareroot(a) thing.

Srikanth Tupurani - 1 year, 10 months ago

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