Prove the integral bound $\int_0^1 \sqrt{\frac{1-x^2}{1+x^2}} d x>\ln 2$

Question

Prove that $$I=\int_0^1 \sqrt{\frac{1-x^2}{1+x^2}} d x>\ln 2$$

My try: $$\begin{aligned} & \forall x \in(0,1), \sqrt{1-x^2}>1-x^2 \\ & \Rightarrow \frac{\sqrt{1-x^2}}{\sqrt{1+x^2}}>\frac{1-x^2}{\sqrt{1+x^2}} \\ & \Rightarrow I>\int_0^1 \frac{\left(1-x^2\right) d x}{\sqrt{1+x^2}} \\ & \Rightarrow I>J-W \end{aligned}$$

Where $J=\int_{0}^{1} \frac{1}{\sqrt{1+x^2}}\:dx,\:W=\int_{0}^{1} \frac{x^2}{\sqrt{1+x^2}}\:dx$

Now we have $$\begin{aligned} & J=\left.\ln \left(x+\sqrt{1+x^2}\right)\right|_0 ^1 \\ & \Rightarrow J=\ln (\sqrt{2}+1) \end{aligned}$$

Also: $$\begin{aligned} W & =\int_0^1 \frac{x^2 d x}{\sqrt{1+x^2}} \\ \Rightarrow W & =\int_0^1 \sqrt{1+x^2} d x-J \\ \Rightarrow W & =\frac{x}{2} \sqrt{1+x^2}+\left.\frac{1}{2} \ln \left(x+\sqrt{x^2+1}\right)\right|_0 ^1-J \\ \Rightarrow W & =\frac{1}{\sqrt{2}}+\frac{1}{2} \ln (\sqrt{2}+1)-\ln (\sqrt{2}+1) \\ \Rightarrow W & =\frac{1}{\sqrt{2}}-\frac{\ln (\sqrt{2}+1)}{2} \end{aligned}$$

Finally $$\begin{aligned} I & >J-W \\ \Rightarrow \quad & I>\frac{3}{2} \ln (\sqrt{2}+1)-\frac{1}{\sqrt{2}}=0.6149 \end{aligned}$$

So I missed to reach the bound of $\ln 2$

Answer 1

Notice that $$\sqrt\frac{1 - x^2}{1 + x^2} = \frac{1 - x^2}{\sqrt{1 - x^4}} \geq (1 - x^2) \left(1 + \frac12 x^4\right),$$ so $$\int_0^1 \sqrt\frac{1 - x^2}{1 + x^2} \,dx \geq \int_0^1 (1 - x^2) \left(1 + \frac12 x^4\right) \,dx = \frac{73}{105} > \frac{25}{36} > \log 2 .$$ The last inequality is well-known and can anyway be derived quickly from an appropriate series.

Answer 2

First Solution. For $0 < a < 1$, we have

\begin{align*} \frac{1}{\sqrt{1-a}} = \int_{0}^{1} \frac{1}{(1 - a y^2)^{3/2}} \, \mathrm{d}y > \int_{0}^{1} \frac{1}{1 - a y^2} \, \mathrm{d}y. \tag{1} \end{align*}

Plugging $\text{(1)}$ to OP's integral $I = \int_{0}^{1} \sqrt{\frac{1-x^2}{1+x^2}} \, \mathrm{d}x$, we get

\begin{align*} I &= \int_{0}^{1} \frac{1-x^2}{\sqrt{1-x^4}} \, \mathrm{d}x \\ &> \int_{0}^{1} \int_{0}^{1} \frac{1-x^2}{1 - x^4 y^2} \, \mathrm{d}y \mathrm{d}x \\ &= \int_{0}^{1} \int_{0}^{x} \frac{1-x^2}{1 - u^4} \cdot \frac{2u}{x^2} \, \mathrm{d}u \mathrm{d}x \tag{$u = x\sqrt{y}$} \\ &= \int_{0}^{1} \int_{u}^{1} \frac{2u(\frac{1}{x^2}-1)}{1 - u^4} \, \mathrm{d}x \mathrm{d}u \tag{Fubini} \\ &= \int_{0}^{1} \frac{2(1-u)^2}{1 - u^4} \, \mathrm{d}u \\ &= \int_{0}^{1} \left( \frac{2}{1+u} - \frac{2u}{1+u^2} \right) \, \mathrm{d}u \tag{partial fractions} \\ &= \log 2. \end{align*}

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Second Answer. We have

\begin{align*} \int_{0}^{1} \sqrt{\frac{1-x^2}{1+x^2}} \, \mathrm{d}x &= \int_{0}^{1} \frac{1-x^2}{\sqrt{1-x^4}} \, \mathrm{d}x \\ &= \int_{0}^{1} \sum_{n=0}^{\infty} \binom{-1/2}{n} (-x^4)^n (1-x^2) \, \mathrm{d}x \\ &= \sum_{n=0}^{\infty} \frac{(2n-1)!!}{(2n)!!} \int_{0}^{1} (x^{4n} - x^{4n+2}) \, \mathrm{d}x \\ &= \sum_{n=0}^{\infty} \frac{(2n-1)!!}{(2n)!!} \frac{2}{(4n+1)(4n+3)} \\ &> \sum_{n=0}^{\infty} \frac{(2n-1)!!}{\color{red}{(2n+1)!!}} \frac{2}{(4n+1)(4n+3)} \tag{3} \\ &= \sum_{n=0}^{\infty} \frac{4}{(4n+1)(4n+2)(4n+3)} \\ &= \log 2. \end{align*}

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Remark. The second solution is essentially the first solution in disguise. Indeed, expanding both sides pf $\text{(1)}$ in $a$ gives

$$ \sum_{n=0}^{\infty} \frac{(2n-1)!!}{(2n)!!} a^n > \sum_{n=0}^{\infty} \frac{1}{2n+1} a^n, $$

which is precisely what we utilized in $\text{(3)}$.

Answer 3

We may use the polynomial lower bounds of the integrand, e.g. the Taylor approximation.

We can directly do the Taylor approximation for the integrand. Alternatively, we can deal with $\frac{1}{\sqrt{1 - x^4}}$ as we can write $$\sqrt{\frac{1-x^2}{1+x^2}} = \frac{1 - x^2}{\sqrt{1 - x^4}}.$$

We have $$\frac{1}{\sqrt{1 - x^4}} \ge 1 + \frac12 x^4.$$ (See the Remarks at the end.)

Thus, we have $$\sqrt{\frac{1-x^2}{1+x^2}} \ge (1 - x^2)(1 + x^4/2)$$ and $$\int_0^1 \sqrt{\frac{1-x^2}{1+x^2}}\, \mathrm{d} x \ge \int_0^1 (1 - x^2)(1 + x^4/2)\, \mathrm{d} x = \frac{73}{105}. $$

We need to prove that $73/105 > \ln 2$. We can use the integral trick. We have $$0 \le \int_0^1 \frac{x^3(1 - x)^3}{1 + x}\,\mathrm{d} x = \frac{111}{20} - 8\ln 2$$ which gives $\ln 2 \le \frac{111}{160} < \frac{73}{105}$.

We are done.

Remarks.

Way 1.

By AM-GM, we have $$\frac{1}{\sqrt{1 - x^4}} + \frac{1}{\sqrt{1 - x^4}} + (1 - x^4) \ge 3$$ which yields $$\frac{1}{\sqrt{1 - x^4}} \ge 1 + \frac12x^4.$$ (Note: This is a trick in Olympiad inequalities. We will see later this is also the Taylor approximation.)

Way 2.

We have the Taylor expansion $$\frac{1}{\sqrt{1 - x^4}}= 1 + \frac12 x^4 + \frac38 x^8 + \cdots$$ We may first try $1 + \frac12 x^4$ and prove it to be a lower bound.

Answer 4

As commented we can use trapezoidal rule if the function in the OP integral is concave denote $f(x)$ be a such function then :

$$f''(x)=(-6x^4+4x^2-2)/(something positive)<0,0<x<1$$

Then use trapezoidal rule.

Proof :

$$I=\int_{0}^{1}\sqrt{\frac{1-x^2}{1+x^2}}dx=\int_{0}^{1}\frac{1}{2}\sqrt{\frac{1-x}{x+x^2}}dx=\int_{0}^{1}\frac{1}{2}\sqrt{\frac{1}{x}-1+\frac{x-1}{x+1}}dx$$

Using a parabola we have :

$$I\ge \int_{0}^{1/3}\frac{1}{2}\sqrt{\frac{1}{x}-1+\frac{x^2-1}{x+1}}dx+\int_{1/3}^{1}\frac{1}{2}\sqrt{\frac{1}{x}-1+\frac{3}{4}\frac{x^2-1}{x+1}}dx$$

Now by direct comparison :

$$I\ge \int_{0}^{1/3}x\sqrt{\frac{1}{x^2}-1+x^2-1}dx+\int_{1/3}^{1}x\sqrt{\frac{1}{x^2}-1+\frac{3}{4}\frac{x^4-1}{x^2+1}}dx>\ln 2$$

Remark : I think it's very classical so the community answer