Question: Show that (x + y)r < xr + yr whenever x and y are positive real numbers and r is a real number with 0 < r < 1.
Solution: Without loss of generality we can assume that x + y = 1 [To see this, suppose we have proved the theorem with the assumption x + y = 1. Suppose that x + y = t. Then x + y = t. Then (x/t) + (y/t) = 1, which implies that ((x/t) + (y/t))r < (x/t)r + (y/t)r. Multiplying both sides of this last equation by tr shows that (x + y)r < xr + yr.]
Assuming that x + y = 1, because x and y are positive, we have 0 < x < 1 and 0 < y < 1. Because 0 < r < 1, it follows that 0 < 1-r < 1, so x1-r < 1 and y1-r < 1. This means that x < xr and y < yr.
Consequently, xr + yr > x + y = 1. This means that (x + y)r = 1r < xr + yr. This proves the theorem for x + y = 1.
Because we could assume x + y = 1, without loss of generality, we know that (x+y)r < xr + yr whenever x and y are positive real numbers and r is a real number with 0 < r < 1.
I am unable to the understand the above mentioned proof. I am also not able to relate the use of without loss of generality in this solution. As far as I understand, when the phrase “without loss of generality” is used in a proof, we assert that by proving one case of a theorem, no additional argument is required to prove other specified cases. I also doubt if the above proof is correct or not.
Note: The above question is an example in given in the book Discrete Mathematics and applications by Kenneth H. Rosen (Indian Adaptation by Kamala Krithivasan).
I wouldn't how to make it more explicit than it is.
You suppose that $x+y=1$ because if it weren't, you can reduce it to this case and move on with the proof. If $x,t$ are any numbers, then $x+y=t$ for some $t$.
To be able to move on with the proof "without loss of generality" we must show that if the theorem is true for the particular case, this is if we have proved it for $x+y=1$, then we can apply it for $x+y=t$, reducing the general case to the particular.
How to go on with this reduction? The author says that if $x+y=t$ then $(x/t)+(y/t)=1$. Now, if it is true that $(a+b)^r<a^r+b^r$ when $a+b=1$ (this is our ASSUMPTION), we can take $a=x/t, b=y/t$, which leave us with $(x/t+y/t)^r<(x/t)^r+(y/t)^r$. The final step: if we multiply both sides by $t^r$ we have $t^r(x/t+y/t)^r<t^r(x^r/t^r)+t^r(y^r/t^r)\implies [t(x/t+y/t)]]^r<t^r(x^r/t^r)+t^r(y^r/t^r)\implies (x+y)^r<x^r+y^r$.
We have proved the general case using the particular case when $x+y=1$, which we have used in our assumption. That particular case needs to be proved, otherwise the whole proof goes astray since it depends of it, and that's what the author does