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anemone
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Prove that $\sqrt{\sin x}>\sin \sqrt{x}$ for $0<x<\dfrac{\pi}{2}$.
anemone said:Prove that $\sqrt{\sin x}>\sin \sqrt{x}$ for $0<x<\dfrac{\pi}{2}$.
castor28 said:We will prove the result separately for the intervals $(0,1]$ and $\left(1,\frac{\pi}{2}\right]$.
Assume first that $1<x<\frac{\pi}{2}$. Since $0<\sin x < 1$, we have $\sqrt{\sin x}>\sin x$. Since $x>1$, we have $x>\sqrt{x}$, and, as $\sin x$ is an increasing function, $\sin x > \sin \sqrt{x}$. We may therefore conclude:
$$\displaystyle
\sqrt{\sin x}>\sin x>\sin \sqrt{x}
$$
in this case.
Assume now that $0<x\leq 1$. This implies that $x<\sqrt{x}$.
Since $\sqrt{x}>0$, the inequality is equivalent to:
$$\displaystyle
\begin{align*}
\frac{\sqrt{\sin x}}{\sqrt{x}} &> \frac{\sin \sqrt{x}}{\sqrt{x}}\\
\sqrt{\frac{\sin x}{x}} &> \frac{\sin \sqrt{x}}{\sqrt{x}} \\
\sqrt{f(x)} &> f(\sqrt{x})
\end{align*}
$$
where $f(x) = \frac{\sin x}{x}$.
Because $0<f(x)<1$ in the interval under consideration, $\sqrt{f(x)} > f(x)$.
Now, the derivative of $f(x)$ is:
$$\displaystyle
\frac{df}{dx} = \frac{x\cdot\cos x - \sin x}{x^2}
$$
and this is negative, since $0< x< \tan x$. This means that $f$ is strictly decreasing; as $x<\sqrt{x}$, $f(x) > f(\sqrt{x})$, and we conclude:
$$\displaystyle
\sqrt{f(x)} > f(x) > f(\sqrt{x})
$$
which completes the proof.
The inequality that needs to be proven is $\sqrt{\sin x} > \sin\sqrt{x}$ for all values of x between 0 and $\frac{\pi}{2}$.
This inequality is important because it is a fundamental result in the study of trigonometric functions and serves as a building block for more complex mathematical proofs. It also has practical applications in fields such as physics and engineering.
The inequality can be proven using various methods such as algebraic manipulation, calculus, and geometric reasoning. One possible approach is to use the Maclaurin series expansion for $\sin x$ and $\sqrt{\sin x}$ and compare their terms.
No, this inequality only holds for values of x between 0 and $\frac{\pi}{2}$. For other values of x, the inequality may not hold or may need to be modified.
Yes, this inequality has practical applications in fields such as signal processing, where it is used to analyze and manipulate signals. It is also used in the study of vibrations and waves in physics and engineering.