# 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 $x>0$.

• 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) – Luis Gimeno Sotelo Jun 3 '17 at 17:59

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

Hint

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

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!