The inequality $\sqrt{\sin x} > \sin \sqrt{x}$ holds true for the interval $0 < x < \frac{\pi}{2}$. This conclusion is supported by analyzing the behavior of both functions within the specified range. The discussion emphasizes the mathematical rigor required to prove this inequality, showcasing the contributions of forum member castor28, who provided insightful reasoning and proof techniques. The proof involves examining the derivatives and concavity of the functions involved.
PREREQUISITESMathematicians, students studying calculus, and anyone interested in advanced inequality proofs will benefit from this discussion.
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.