Proving an inequality using calculus I was going over a solution recently, which lays out a subtle argument. I believe that it is not rigorous enough.
The inequality is given below:-
$\cos(x)\gt \frac{1}{1+x^2}$ for all $x$ belonging to the interval $[0,\frac{\pi}{4}]$.
The solution argued that the concavity of both the functions is opposite, that is one function is concave up whereas the other function is concave down.
However I was not satisfied by this approach. Could anyone give a better approach or explain this argument more rigorously.
Thank You.
 A: We can establish the inequality without appealing to calculus.  Recall from geometry that $\sin x\le x$ for $0\le x\le \pi/2$.  
This inequality is equivalent to the inequalities
$$\begin{align}
\sin^2x&\le x^2\\\\
1-\cos^2x&\le x^2\\\\
\cos^2x&\ge 1-x^2\tag 1\\\\
\end{align}$$
For $0\le x\le 1$, we may take the square root of both sides of $(1)$ and write
$$\cos x\ge \sqrt{1-x^2}$$
Next, it is easy to show that for $0\le x\le \sqrt{\frac{\sqrt{5}-1}{2}}>\frac{\pi}{4}$, $\sqrt{1-x^2}\ge \frac{1}{1+x^2}$.  This is left as an exercise.  
Therefore, for $0\le x\le \frac{\pi}{4}$, we have the desired inequality 
$$\cos x\ge \frac{1}{1+x^2}$$

Using Calculus 
From the extended law of the mean, there exists a number $0<\xi<x$ such that $\cos x=1-\frac12 \cos (\xi)x^2$.  
Since $|\cos \xi|\le 1$, we see that $\cos x\ge 1-\frac12 x^2$.  
It is easy to show that for $|x|\le 1$, $1-\frac12 x^2\ge \frac{1}{1+x^2}$, whence we have the desired inequality again!

SIDE NOTE:
As an interesting side note, we also have from geometry the inequality $\sin x\ge x\cos x$ for $0\le x\le \pi/2$.  From this, it is easy to show that 
$$\cos x\le \frac{1}{\sqrt{1+x^2}}$$
A: Here's another way to get the inequality without using any calculus.
For clarity of exposition, it is convenient to rewrite the desired inequality as $(1+\theta^2)\cos\theta\gt1$ for $0\lt\theta\lt\pi/4$.
Draw the unit circle and draw an angle $\theta$ in the first quadrant, so that $0\lt\theta\lt\pi/2$.  Measured in radians, $\theta$ is the arc length along the circle from $(1,0)$ to $(\cos\theta,\sin\theta)$, which is greater than the length of the chord connecting the same two points.  This inequality can be written as
$$\theta^2\gt(1-\cos\theta)^2+(\sin\theta)^2$$
Expanding this and using the trig identity $\sin^2\theta+\cos^2\theta=1$, we can conclude
$$1+\theta^2\gt3-2\cos\theta$$
for all $0\lt\theta\lt\pi/2$.  Hence to prove the desired inequality, it suffices to show that $(3-2\cos\theta)\cos\theta\gt1$ for $0\lt\theta\le\pi/4$.  But this can be rewritten as $0\gt1-3\cos\theta+2\cos^2\theta$ or, in factored form, as
$$0\gt(1-\cos\theta)(1-2\cos\theta)$$
This is clearly satisfied when $1\gt\cos\theta\gt1/2$, which is to say when $0\lt\theta\lt\pi/3$.  Since $\pi/4\lt\pi/3$, we have the desired inequality.
A: I am tempted to approach this problem by using what's called the "Racetrack principle". It can be proved using the Mean Value Theorem. Here's a link: https://en.wikipedia.org/wiki/Racetrack_principle
Statement: Let $f,g$ be continuously differentiable functions on $\mathbb{R}$.If, $f(0)=g(0)$ and $f^{'}(x)>g^{'}(x)$ for all $x \geq 0$ then $f(x)>g(x)$ for all $x \geq 0$.
A proof of this result can be written by taking $h=f-g$ and applying the Mean Value Theorem to it.
Before we proceed, I hope you're familiar with the following inequality: $Sin(x)<x$ for all $x \in \mathbb{R}$.
Take $f(x)=Cos(x)$ and $g(x)=\frac{1}{1+x^{2}}$ in $[0,\frac{\pi}{4}]$. $f(0)=g(0)=1$. What we now need to show is that the function $H(x) = f(x)-g(x)$ has positive derivative.
We get $f^{'}(x)= -Sin(x)$ and $g^{'}(x)= \frac{-2x}{1+x^{2}}$. So,
\begin{equation}
Sin(x)<x \implies -Sin(x)>-x \\
\implies H^{'}(x)=\frac{2x}{1+x^{2}}-Sin(x)>\frac{2x}{1+x^{2}}-x\\
\end{equation}
On $[0,\frac{\pi}{4}]$ , $x(\frac{2}{1+x^{2}}-x)>0$. So, $H^{'}(x)>0$ and we can apply the result to show the desired inequality.
