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

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### 1 Answer

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**