0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I don´t know how to use it here because I don´t know how to express "ln(x)" in terms of "x^2" (I have tried Taylor, but the question is about using MVT) $\endgroup$ – Luis Gimeno Sotelo Jun 3 '17 at 17:59
1
$\begingroup$

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 $$f (x)\le 0.$$

$\endgroup$
0
$\begingroup$

Hint

Use MVT with $f(t)=\ln(t)$ on $[1,x^2]$. You have to separate the cases $x>1$ and $x\in ]0,1[$.

$\endgroup$
0
$\begingroup$

We need to prove that $$\ln{x}\leq\frac{x^2-1}{2}.$$ Let $0<x\leq1$ and $x=\frac{1}{y}$.

Thus, $y\geq1$ and we need to prove that $$\ln\frac{1}{y}\leq\frac{\frac{1}{y^2}-1}{2}$$ or $$\ln{y}\geq\frac{y^2-1}{2y^2}.$$ Indeed, for $y=1$ it's obvious. Let $y>1$. Thus, there is $1<\theta\leq y$, for which $$\frac{\ln{y}-\ln1}{y-1}=\frac{1}{\theta},$$ which says that we need to prove that $$\frac{y-1}{\theta}\geq\frac{y^2-1}{2y^2}$$ or $$2y^2\geq(y+1)\theta$$ and since $\theta\leq y$, it's enough to prove that $$2y^2\geq(y+1)y,$$ which is obvious.

Let $x>1$. Hence, there is $1<\theta\leq x$ for which $$\frac{\ln{x}-\ln1}{x-1}=\frac{1}{\theta}.$$

Thus, we need to prove that $$\frac{x-1}{\theta}\leq\frac{x^2-1}{2}$$ or $$\theta(x+1)\geq2,$$ which is obvious ($\theta\geq1$ and $x\geq1$).

Done!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.