Thursday, 1 January 2015

calculus - Prove inequality using Mean Value Theorem




How would you prove the following inequality using the Mean Value Theorem:
$$1+2\ln x\leq x^2$$ for $x>0$.



Answer



Let $f (t)=1+2\ln(t)-t^2$ for $t>0$.




for $x>0$ , $f $ is continuous at $[x,1] $ or $[1,x]$ and differentiable at $(x,1) $ or $(1,x) $ thus by MVT, exists $c$ strictly between $x $ and $1$ such that



$$f (x)-f (1)=(x-1)f'(c) $$
$$1+2\ln (x)-x^2=2(x-1)(1/c -c) $$
$$=2 (x-1)\frac {1-c^2}{c} $$



If $x>1$ then $1-c^2<0$ and



if $x <1$ then $1-c^2>0$




thus in all cases
$$f (x)\le 0.$$


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...