The Schwarz inequality
is
\[
\left( \int_a^b \f(x)\, \g(x)\,\d x\right)^{\!\!2}
\le
\left(
\int_a^b \big( \f(x)\big)^2 \d x
\right)
\left(
\int_a^b \big( \g(x)\big)^2 \d x
\right)
.
\tag{\(*\)}
\]
By setting \( \f(x)=1\) in \((*)\), and choosing \(\g(x)\), \(a\) and \(b\) suitably, show that for \(t> 0\,\),
\[
\frac {\e^t -1}{\e^t+1} \le \frac t 2
\,.
\]
By setting \( \f(x)= x\) in \((*)\), and choosing \( \g(x)\) suitably, show that
\[
\int_0^1\e^{-\frac12 x^2}\d x \ge 12 \big(1-\e^{-\frac14})^2
\,.
\]
Use \((*)\) to show that
\[
\frac {64}{25\pi} \le \int_0^{\frac12\pi}
\!\!
{\textstyle \sqrt{\, \sin x\, } }
\, \d x
\le \sqrt{\frac \pi 2 }
\,.
\]
Solution:
Let \(f(x) = 1, g(x) = e^x, a = 0, b = t\), so
\begin{align*}
&& \left ( \int_0^t e^x \d x \right)^2 &\leq \left (\int_0^t 1^2 \d x \right) \cdot \left (\int_0^t (e^x)^2 \d x \right) \\
\Rightarrow && (e^t-1)^2 &\leq t \cdot (\frac12e^{2t} - \frac12) \\
\Rightarrow && \frac{e^t-1}{e^t+1} & \leq \frac{t}{2}
\end{align*}
Let \(f(x) = x, g(x) = e^{-\frac14 x^2}, a = 0, b = 1\)
\begin{align*}
&& \left ( \int_0^1 xe^{-\frac14 x^2} \d x \right)^2 &\leq \left (\int_0^1 x^2 \d x \right) \cdot \left (\int_0^1 (e^{-\frac14x^2})^2 \d x \right) \\
\Rightarrow && \left ( \left [-2e^{-\frac14x^2} \right]_0^1 \right)^2 & \leq \frac{1}{3} \int_0^1 e^{-\frac12 x^2} \d x \\
\Rightarrow && \int_0^1 e^{-\frac12 x^2} \d x & \geq 12(1-e^{-\frac14})^2
\end{align*}
Let \(f(x) = 1, g(x) = \sqrt{\sin x}, a = 0, b = \tfrac12 \pi\), then
\begin{align*}
&& \left ( \int_0^{\frac12 \pi} \sqrt{\sin x} \d x \right)^2 &\leq \left (\int_0^{\frac12 \pi} 1^2 \d x \right) \cdot \left (\int_0^{\frac12 \pi}|\sin x| \d x \right) \\
&&&= \frac{\pi}{2} \cdot 1 \\
\Rightarrow && \int_0^{\frac12 \pi} \sqrt{\sin x} \d x & \leq \sqrt{\frac{\pi}{2}}
\end{align*}
Let \(f(x) =(\sin x)^{\frac14}, g(x) = \cos x, a = 0, b = \tfrac12 \pi\), so
\begin{align*}
&& \left ( \int_0^{\frac12 \pi} (\sin x)^{\frac14} \cos x \d x \right)^2 &\leq \left (\int_0^{\frac12 \pi} \cos^2 x \d x \right) \cdot \left (\int_0^{\frac12 \pi}\sqrt{\sin x} \d x \right) \\
\Rightarrow &&\left ( \left [\frac45 (\sin x)^{\frac54} \right]_0^{\frac12 \pi} \right)^2 & \leq \frac{\pi}{4} \int_0^{\frac12 \pi}\sqrt{\sin x} \d x \\
\Rightarrow && \frac{64}{25\pi} &\leq \int_0^{\frac12 \pi}\sqrt{\sin x} \d x
\end{align*}