
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
January 7th, 2018, 05:05 AM  #1 
Newbie Joined: Jan 2018 From: Czechia Posts: 1 Thanks: 0  Prove inequality for every positive real number
Hi, I have a problem which goes like this: Prove that for every positive real number t the following is valid: However, I have no idea how to solve it. Could anyone help me please? 
January 7th, 2018, 02:30 PM  #2 
Senior Member Joined: Oct 2009 Posts: 456 Thanks: 149 
Let's split it up in three subproblems: First, prove for all $t\geq 0$ that $$\frac{t^2 + 1}{t+1}\geq \sqrt{t}$$ Second, prove that for $0\leq t\leq 1$, that $$\frac{t^2 + 1}{t+1}  \sqrt{t} \leq 1t$$ Finally, prove that for $t\geq 1$, that $$\frac{t^2 + 1}{t+1}  \sqrt{t} \leq t1$$ Several ways of doing this. You could isolate the square root and then square both sides (but be careful with doing this, as you don't want to square negative numbers: 2<1, but 4<1 is not true). 
January 8th, 2018, 05:49 PM  #3  
Senior Member Joined: May 2016 From: USA Posts: 1,125 Thanks: 467  Quote:
$\text {CASE I: } 0 < t < 1 \implies 0 < t^3 < t^2 < t < \sqrt{t} < 1.$ $0 < t < 1 \implies \ 1 < t  1 < 0 \implies 0 < t  1 < 1 \implies$ $t  1 = 1  t \text { and } 0 < 1  t < 1.$ $0 < t^2 < t \implies 0 < t^2 + 1 < t + 1 \implies 0 < \dfrac{t^2 + 1}{t + 1} < 1 \implies 0 < \dfrac{t^2 + 1}{t + 1}  t < 1  t.$ $0 < t < \sqrt{t} \implies \ \sqrt{t} < \ t \implies \dfrac{t^2 + 1}{t + 1}  \sqrt{t} < \dfrac{t^2 + 1}{t + 1}  t < 1  t \implies \dfrac{t^2 + 1}{t + 1}  \sqrt{t} \le t  1.$ $0 < t^3 < t^2 < t < 1 \implies 0 < t^3 < 1 \text { and } 0 < 1  t \implies$ $(1  t) * 0 < (1  t) * t^3 < (1  t) * 1 \implies 0 < t^3  t^4 < 1  t \implies$ $t^3 + t < t^4 + 1 \implies t^3 + 2t^2 + t < t^4 + 2t^2 + 1 \implies$ $t(t^2 + 2t + 1) < t^4 + 2t^2 + 1 \implies t(t + 1)^2 < (t^2 + 1)^2 \implies$ $t < \dfrac{(t^2 + 1)^2}{(t + 1)^2} \implies \sqrt{t} < \dfrac{t^2 + 1}{t + 1} \implies 0 < \dfrac{t^2 + 1}{t + 1}  \sqrt{t}.$ $\therefore 0 < t < 1 \implies 0 \le \dfrac{t^2 + 1}{t + 1}  \sqrt{t} \le t  1.$ $\text {CASE II: } 0 < 1 \le t.$ You can take it from here.  

Tags 
inequality, number, numbers, positive, prove, real 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
real number x for which inequality  jemes  Algebra  4  November 26th, 2014 02:58 PM 
USE REAL NUMBER AXIOMS TO PROVE THAT (1).(1)=1 Urgent  th3j35t3r  Real Analysis  2  November 17th, 2013 07:53 AM 
If n is a positive odd number, prove that 24 divides n³  n  usermind  Applied Math  11  May 28th, 2012 06:32 AM 
Positive Real Numbers  jstarks4444  Real Analysis  1  March 30th, 2011 04:38 AM 
Positive Real Numbers  jstarks4444  Number Theory  1  December 31st, 1969 04:00 PM 