# An Old Inequality

### Solution 1

We may rewrite the inequality as $\displaystyle 1-\frac{(a-b)^2}{(a+b)^2}\ge\cos\left(\frac{\pi}{2}\cdot\frac{a-b}{a+b}\right)$ which suggests that the problem may be reduced to a study of the function $\displaystyle f(x)=1-x^2-\cos\frac{\pi x}{2}.\,$ The function is even; suffice it to consider $x\ge 0.\,$ Additionally, since both $a\,$ and $b\,$ are assumed to be positive, which makes $\displaystyle \left|\frac{a-b}{a+b}\right|\le 1\,$ (even when one or both $a,b\to 0),\,$ we may restrict our interest to $x\le 1.\,$ The plan is to show that, for $x\in [0,1],\,$ $f(x)\ge 0.$

Thus, we have $\displaystyle f'(x)=-2x+\frac{\pi}{2}\sin\frac{\pi x}{2}\,$ and $\displaystyle f''(x)=-2+\frac{\pi^2}{4}\cos\frac{\pi x}{2}.$

Let's remark that $f''\,$ is strictly decreasing and, since $\displaystyle f''(0)=-2+\frac{\pi^2}{4}\gt 0,\,$ whereas $f''(1)=-2\lt 0,\,$ we deduce that for some $t\in (0,1),\,$ $f'\,$ is strictly increasing on $[0,t]\,$ and strictly decreasing on $[t,1].$

But $f'(0)=0\,$ and $\displaystyle f'(1)=-2+\frac{\pi}{2}\lt 0;\,$ hence, for some $u\in (t,1),\,$ $f\,$ is strictly increasing on $[0,u]\,$ and decreasing on $[u,1].\,$ It follows that $\min_{x\in [0,1]}f(x)=\min\{f(0),f(1)\}=0.\,$

### Solution 2

As $\cos x$ is an even function of $x$, we can assume $a\geq b$ without loss in generality. Let $k=a/b$. Thus, with $k\geq 1$, the inequality can be written as

$\displaystyle \frac{4k}{(k+1)^2}\geq \cos\left[\frac{\pi}{2}\frac{(k-1)}{(k+1)}\right].$

Let,

$\displaystyle x=\frac{\pi}{4}\frac{(k-1)}{(k+1)}.$

Thus,

$RHS=\cos 2x = 1-2\sin^2 x,$

and the inequality becomes,

\displaystyle \begin{align} 1-\left(\frac{4x}{\pi}\right)^2&\geq 1-2\sin^2 x \\ \sin^2 x &\geq \left(\frac{2\sqrt{2}x}{\pi}\right)^2 \end{align}

Noting that $0\leq x\leq \pi/4$, the inequality is equivalent to

$\displaystyle \sin x \geq \frac{2\sqrt{2}x}{\pi}.$

$\sin x$ is a concave function in the first octant. Thus, applying Jensen's inequality, $\forall \alpha\in[0,1]$,

$\displaystyle \sin\left[\alpha\frac{\pi}{4}+(1-\alpha)0\right]\geq\alpha\sin\frac{\pi}{4}=\frac{\alpha}{\sqrt{2}}.$

Plugging in $\alpha=4x/\pi$, we get the desired inequality.

### Acknowledgment

Leo Giugiuc has kindly posted the problem at the CutTheKnotMath facebook page, along with a solution of his (Solution 1). Solution 2 is by Amit Itagi. Leo also included a link to the original post by Abhay Chandra at the mathematical inequalities facebook group. The inequality is well known, although I could not pinpoint a suitable reference.