Minimum value

Algebra Level 5

If a , b , c , a,b,c, and d d are positive real numbers, what is the infimum (minimum) value of

a b + c + d + b a + c + d + c a + b + d + d a + b + c ? \sqrt { \frac { a }{ b+c+d } } +\sqrt { \frac { b }{ a+c+d } } +\sqrt { \frac { c }{ a+b+d } } +\sqrt { \frac { d }{ a+b+c } } ?

Round your answer to the nearest hundredth.


The answer is 2.00.

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1 solution

the minimum value of 2 of the terms can be 0 so let a=0 b=0 we need to find the minimum value of (c/d)^1/2 + (d/c)^1/2=2

Moderator note:

Wrong, 0 is not a positive number.

Edit: Oh, good catch, Calvin. None of a , b , c , d a,b,c,d can be 0 0 . Is this problem completely botched now? Because, we could pick c , d c,d as arbitrarily close to 0 0 , and get an minimum as arbitrarily close to 2 2 . Maybe the problem should be reworded to say a , b , c , d 0 a,b,c,d \ge 0 , but the minimum has to be determinate?

Edit 2: Problem now slightly re-worded.

(Ignore the following until this is resolved.)

Here's a non-rigorous approach. First, we note that if we consider "lines" passing through the origin { 0 , 0 , 0 , 0 } \left\{ 0,0,0,0 \right\} , using a a as a parameter, that is

a = a b = a B c = a C d = a D a=a\\ b=aB\\ c=aC\\ d=aD

then the expression becomes a constant depending only on B , C , D B,C,D , that is

1 B + C + D + B C + D + 1 + C D + 1 + B + D 1 + B + C \sqrt { \frac { 1 }{ B+C+D } } +\sqrt { \frac { B }{ C+D+1 } } +\sqrt { \frac { C }{ D+1+B } } +\sqrt { \frac { D }{ 1+B+C } }

To minimize this (this is the hand waving part), we simply let C = 0 C=0 and D = 0 D=0 , and we're left with

1 B + B \sqrt { \frac { 1 }{ B } } +\sqrt { B }

to minimize. A little bit of calculus or other gets us the minimum of 2 2 .

The salient point of this approach is that if we were to attempt using partial differentials, we won't find any local maxima or minima because the value of the expression is a constant along any straight line through the origin. Hence, we have to look for maxima or minima along boundaries or where some of a , b , c , d a,b,c,d could be 0 0 , but not all of them, as the value is indeterminate at origin.

This could be kind of a hard problem to visualize because it's really 4D geometry.

Michael Mendrin - 6 years ago

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The idea is more of "find the infimum", and it occurs when a , b 0 , c = d a, b \rightarrow 0, c = d .

I have edited the problem accordingly.

Calvin Lin Staff - 6 years ago

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Shee, I had to go look that up, "infimum". But that's right on target for this problem.

Michael Mendrin - 6 years ago

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@Michael Mendrin Yea, it's a calculus term, which is why I almost always add "(minimum)" after it. The "better" way of posing this problem would be to restrict/normalize it to a + b + c + d = 1 a+b+c+d = 1 .

Calvin Lin Staff - 6 years ago

I believe this problem can be solved using simple inequalities but I couldn't finish it off.

WLOG assume a b c d a\leq b \leq c \leq d with a + b + c + d = 1 a+b+c+d=1 , and denote a , b , c , d a,b,c,d as sin 2 W , sin 2 X , sin 2 Y , sin 2 Z \sin^2 W, \sin^2 X, \sin^2 Y, \sin^2 Z respectively for 0 < W X Y Z π 2 0<W \leq X \leq Y\leq Z\leq \frac{\pi}2 .

We have sin 2 W + sin 2 X + sin 2 Y + sin 2 Z = 1 \sin^2 W+ \sin^2 X + \sin^2 Y + \sin^2 Z = 1 and we want to prove that tan W + tan X + tan Y + tan Z 4 3 \tan W + \tan X + \tan Y + \tan Z \geq \frac4{\sqrt3} .

Pi Han Goh - 6 years ago

not a very good approach but let d and c be veeeeerrrrrryyyy small

Arijit ghosh Dastidar - 5 years, 12 months ago

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