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#1 2019-02-12 20:05:37

Dr Francis Hung
Guest

Really easy inequality question

Define f(a,b,c) = sqrt(a)+sqrt(b)-sqrt(c)
g(a,b,c) = sqrt(a+b-c)
Prove g(a,b,c)/f(a,b,c)+g(b,c,a)/f(b,c,a)+g(c,a,b)/f(c,a,b)<=3

#2 2019-02-13 01:34:24

Grantingriver
Member
Registered: 2016-02-01
Posts: 122

Re: Really easy inequality question

Hi Dr. Francis Hung, this can be proven easily by observing that a,b and c are greater than or equal to zero since the square root of a negative number is not a real number and the inequality emphasizes that a,b and c are in fact real numbers. However, since the three terms in the left side of the given inequality are identical it is sufficient to prove that each term is less than or equal to 1 hence the prove goes as follows:


remember that

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