Easy Inequality
If a + b = 1 . Then the minimum value of ( a + a 1 ) 2 + ( b + b 1 ) 2 is of the form " m / n " . Where m and n are coprime. Then the value of m + n ?
The answer is 27.
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2 solutions
Since ab<=1/4, if we substitute ab=1/4, then 1/4 is the maximum value of ab, so how can it be the minimum value of the question?
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You seem to be claiming "The minimum value of a function cannot occur at the maximum value of its domain".
This is not true. For example, if we consider the function f ( x ) = 1 − x on the domain [ 0 , 1 ] , then the minimum value of f ( x ) is clearly f ( 1 ) = 0 .
by inspection a = b = 1/2 and a+ 1/a = 5/2 = b+1/b and so (a+ 1/a)^2 + (b+1/b)^2 = 25/2= m/n so m+n = 25+2 = 27
Why must we have a = b = 1 / 2 ? Why can't we have a = 0 . 2 , b = 0 . 8 ?
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Maybe because the expression is symmetric for a and b ?
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Check out inequalities with strange equality conditions for many reasons why this is not true.
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@Calvin Lin – So inequalities are not that easy too!
I know that many times it works but where is the proof.
(a+1/a)^2 +(b+1/b)^2 = a^2+1/a^2 + b^2 +1/b^2 +4 =(a+b)^2 -2ab +(1/a +1/b)^2 - 2/ab +4
by AM- GM, ab <=(a+b /2)^2=1/4 hence we get (a+1/a)^2 +(b+1/b)^2 >= (1-1/2)(1+16) +4 =25/2 so, m+n =25+2=27