# Simple proof help?

simple proof help??

## Homework Statement

The question states: Let rεQ+ Prove that if (r^2+1)/(r)≤1, then (r^2+2)/(r)≤2.
I wanted to prove it trivially by proving it is true for all Q(x). would this be a correct way?

## The Attempt at a Solution

Since (r^2+2)/(r)≤2= (r-1)^2+1≤ 0 it follows (r-1)^2≤-1 for all rεQ+. Therefore (r^2+2)/(r)≤2.

Ray Vickson
Homework Helper
Dearly Missed

The minimum of the function f(r) = (r^2 + 1)/r = r + (1/r) is f_min = 2, at r = 1. So, it is impossible to have f(r) <= 1 with r > 0.

RGV

I realized my mistake you cant prove it trivially. The way I proved it is wrong since (r-1)^2+1≥0 and not ≤0.

why am i getting different graphs when I graph r-(1\r)≤1 and r^2-r+1? I just multiplied through by r and moved everything around but still im getting different equations.

HallsofIvy
Homework Helper

why am i getting different graphs when I graph r-(1\r)≤1 and r^2-r+1? I just multiplied through by r and moved everything around but still im getting different equations.
What do you mean by "r-(1\r)≤1"? there is no "≤" in your second formula. If you dif mean r- (1/r)≤ 1, multiplying through by (positive) r gives r^2- 1≤ r and then "moving everything around" gives r^2- r- 1≤ 0, not r^2- r+ 1≤ 0.

for the r- (1/r)≤ 1 i get a hyperbola looking graph and r^2- r- 1≤ 0 I get a parobla with both being shaded.

Ray Vickson
Homework Helper
Dearly Missed

for the r- (1/r)≤ 1 i get a hyperbola looking graph and r^2- r- 1≤ 0 I get a parobla with both being shaded.
So, are we to understand that you really meant (r^2 - 1)/r ≤ 1? You never told us that.

Anyway, assuming that is the case, the inequalities 1 ≥ (r^2-1)/r = r - 1/r and r > 0 give a region along the real r-axis that you can find by plotting y = r - 1/r and seeing where the graph has y ≤ 1. It is true that the graph is like a hyerbola. As you say, you can also multiply through by r and get another inequality r^2 - r ≤ 1, so we can plot y = r^2 - r and see where y ≤ 1. As you rightly say, the graph is a parabola. However, the points r that satisfy these inequalities are exactly the same: both give an interval (0,r0), with the same r0.

RGV