Homework Help

Prove that if a^2+b^2=1, a>0,b>0, then alna+blnb+(a+b)ln(a+b)<0.

user profile pic

hahaz | Student, Grade 10 | eNoter

Posted May 29, 2009 at 11:51 PM via web

dislike 1 like

Prove that if a^2+b^2=1, a>0,b>0,

then alna+blnb+(a+b)ln(a+b)<0.

Tagged with identity, math

1 Answer | Add Yours

user profile pic

giorgiana1976 | College Teacher | Valedictorian

Posted May 30, 2009 at 9:58 PM (Answer #1)

dislike 0 like

First of all, we'll rearrange the expression given:

a lna +b lnb<-(a+b)*ln(a+b)

We'll divide the inequality above, with the quantity (a+b) and the inequality will become:

[a/(a+b)]*lna + [b/(a+b)]*lnb<-ln (a+b)

If we'll analyze the function f(x)=lnx, we'll see , after calculating the second derivative, that f(x) is a concave function.

f'(x)=1/x, f"(x)=-1/x^2<0, so f(x) is concave.

If f(x) is concave, then we could write the inequality

ln{ [a/(a+b)]*a +  [b/(a+b)]*b}>[a/(a+b)]*lna + [b/(a+b)]*lnb

ln[(a^2 + b^2)/(a+b)]>[a/(a+b)]*lna + [b/(a+b)]*lnb

But a^2 + b^2=1, from hypothesis, so then

ln[1/(a+b)]>[a/(a+b)]*lna + [b/(a+b)]*lnb

ln[(a+b)^-1]>[a/(a+b)]*lna + [b/(a+b)]*lnb

-ln(a+b)]>[a/(a+b)]*lna + [b/(a+b)]*lnb q.e.d

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes