x+y+z=1.

Demonstrate that x/(x+1) + y/(y+1) + z/(z+1)<3/4.

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We are considering the function f(x)=t/(t+1)

f:[0,infinity)->R

We shall calculate the first and the second derivative, in order to see if the function is concave, so that we could apply the Jensen's inequality.

f'(x)=[t'*(t+1)-t*(t+1)']/(t+1)^2

f'(x)=(t+1-t)/(t+1)^2, we'll reduce the similar terms, and the result will be:

f'(x)=1/(t+1)^2

f"(x)=[1'*(t+1)-1*(t+1)']/(t+1)^4

f"(x)=-1/(t+1)^4<0

The denominator, will be always positive; (t+1)^4>0

f"(x) is negative, so the graph of the function is concave.

In this case, we can apply Jensen's inequality.

f[(x+y+z)/3]>[f(x) +f(y)+f(z)]/3

But, from hypothesis, x+y+z=1, so f[(x+y+z)/3]=f(1/3)

f(1/3)=1/3/(1/3 + 1)=1/3/4/3=1/4

f(x) +f(y)+f(z)=x/(x+1) + y/(y+1) + z/(z+1)

[f(x) +f(y)+f(z)]/3=[x/(x+1) + y/(y+1) + z/(z+1)]/3

From Jensen's inequality

[x/(x+1) + y/(y+1) + z/(z+1)]/3<1/4

Multiplying the means,

[x/(x+1) + y/(y+1) + z/(z+1)]<3/4 q.e.d

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