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Starting from $0<a_1\le b_1$, replace the lower endpoint by the geometric mean and the upper endpoint by the arithmetic mean. The arithmetic-geometric mean inequality gives $a_n\le a_{n+1}\le b_{n+1}\le b_n$. The bounded monotone sequence theorem gives limits, and the arithmetic recurrence forces them to coincide. The limiting value is the arithmetic-geometric mean of the initial pair.

An equilateral primordial bispectrum configuration has equal external wavenumbers. For a cubic time-derivative curvature bispectrum, $S\propto k_1{k}_2{k}_3/(k_1+k_2+k_3{)}^3$ is maximized here at fixed perimeter by the arithmetic-geometric mean inequality.

The geometric mean of finitely many nonnegative real numbers is the nonnegative $n$th root of their product. For two positive numbers it is $\sqrt{ab}$. The arithmetic-geometric mean inequality compares it with their arithmetic mean. Unlike an arithmetic average, the geometric mean respects multiplicative scaling and is natural for growth factors.

For a normalized state with the required position and momentum moments, $[x,p]=i\hslash$ and the Robertson uncertainty principle give $\Delta x\Delta p\ge \hslash /2$. In a quantum harmonic oscillator, the arithmetic-geometric mean inequality then bounds the variance contribution to energy below by $\hslash \omega /2$.

Mathematical optimization studies extrema under constraints.

The arithmetic-geometric mean inequality givesBy induction these inequalities hold at every step, so $(a_n)$ is increasing and bounded above by $b$, while $(b_n)$ is decreasing and bounded below by $a$. The bounded monotone sequence theorem gives limits $\alpha ,\beta$ with $0<a\le \alpha \le \beta \le b$. Taking limits in the arithmetic mean recurrence gives $\beta =(\alpha +\beta )/2$, whence $\boxed{\alpha =\beta }$. Thus both sequences converge to the same positive limit, the arithmetic-geometric mean iteration's limit. No elementary closed formula for that limit is needed.

For a normalized state and a self-adjoint operator $Q$ with finite second moment, the quantum uncertainty is defined by $\Delta Q=\Vert (Q-⟨Q⟩)\psi \Vert$. Since $⟨Q⟩$ is real,In an energy eigenstate of the quantum harmonic oscillator,Discarding the nonnegative squared means proves the stated lower bound. The Heisenberg uncertainty principle is $\Delta x\Delta p\ge \hslash /2$, following from $[x,p]=i\hslash$. Applying the arithmetic-geometric mean inequality to the two variance contributions givesHere $\omega >0$ is the oscillator frequency. The bound is the ground-state energy; equality requires zero means and the appropriate minimum-uncertainty width.

For independent and identically distributed random variables sampled from a probability density function $f$, a kernel density estimator with kernel for density estimation $K$ and smoothing bandwidth $h>0$ isThe usual kernel for density estimation is nonnegative; finite $\int K^2$ makes the integrated variance finite. The mean integrated squared error and its bias-variance decomposition of mean squared error areThe interchange is justified by the Tonelli theorem.

For the exponential distribution and the specified unit-width uniform kernel, put $g_h(x)=\mathbb{E}{\hat{f}}_h(x)$. The expected value is $h^{-1}\mathbb{P}\{x-h/2\le X\le x+h/2\}$. Intersecting this interval with the positive half-line givesOn $[-h/4,0)$ the true probability density function vanishes, whereasHere $1-e^{-z}\ge z/2$ for $0\le z\le 1/4$. This leakage across the support boundary proves the integrated squared bias from a density jump:In particular, $h_0=1$ and $c=1/256$ suffice. The phenomenon differs from a smooth interior bias of a kernel density estimator: the order-one boundary error persists over a region of width proportional to $h$.

By independence, $\mathrm{Var}({\hat{f}}_h(x))=n^{-1}\{\mathbb{E}{K}_h(x-X{)}^2-g_h(x{)}^2\}$. Integrating the first term and using $\int K^2=1$ yieldsAlso $I(h)\le \int f^2=1/2$: the kernel density estimator mean is an average of translates of $f$, so the Jensen inequality followed by the Tonelli theorem bounds its squared integral.

To optimize over every $h>0$, we must establish that useful smoothing bandwidths approach zero, rather than optimize only a formal small-$h$ expression. An exact calculation supplies that justification. The autocorrelation integral of the exponential distribution density isAveraging over the two uniform windows, or integrating the piecewise expression for $g_h$, givesThese formulas show that $B$ is continuous on $(0,\infty )$, is strictly positive there because $g_h>0$ on $(-h/2,0)$, and tends to $1/2$ as $h\to \infty$. Consequently, $B$ is bounded away from zero on $[\delta ,\infty )$ for every $\delta >0$. Its expansion agrees with the supplied $B(h)=h/12+o(h)$.

The choice $h_n=\sqrt{12/n}$ givesChoose smoothing bandwidths within $1/n$ of the infimum. Their mean integrated squared error tends to zero, and nonnegativity of the integrated variance gives $B(h_n)\to 0$, hence $h_n\to 0$. For any $\epsilon \in (0,1/12)$, eventuallyThe last step is the arithmetic-geometric mean inequality. Taking the lower limit and then $\epsilon \downarrow 0$ matches the upper bound. ThusThe original PDF has this constant; the TeX aid's $1/\sqrt{2}$ is a transcription error.

Write ${\lambda }_n$ for Lebesgue measure on ${\mathbb{R}}^n$. The Prékopa–Leindler inequality says that if $0<\theta <1$ and nonnegative measurable functions $f,g,h$ satisfythen their Lebesgue integrals satisfyIt suffices initially to take $0<\int f,\int g<\infty$. Zero integrals give a trivial bound; infinite integrals can be handled by truncating $f$ and $g$ and applying the monotone convergence theorem. As usual, a zero factor makes the asserted lower bound zero.

Here is a one-dimensional proof using quantile functions. Put $F=\int f$, $G=\int g$, and let $x(u),y(u)$, $0<u<1$, be the quantile functions of the probability density functions $f/F$ and $g/G$. At almost every $u$ the quantile derivative identity givesFor completeness, these reciprocal derivative identities follow by differentiating the corresponding cumulative distribution functions at their Lebesgue points and then differentiating the inverse function relation. Sampling a quantile function at a uniform argument samples the density itself, so the exceptional set, including locations where the density vanishes or is infinite, has probability zero. Gaps in a density's support can give jumps in its quantile function; they do not invalidate the argument below.

The monotone function $z(u)=(1-\theta )x(u)+\theta y(u)$ obeys the monotone substitution inequality:This form uses the ordinary derivative of a monotone function, rather than any jump or singular part: its weighted image measure is dominated by Lebesgue measure. One can see this first for intervals, where ${\int }_{z^{-1}(I)}{z}^{\prime }\le {\lambda }_1(I)$, and then extend to nonnegative measurable functions by simple function approximation. Thus no assumption of strictly positive smooth probability density functions is hidden in the proof.

Using the hypothesis and the weighted arithmetic-geometric mean inequality, we obtainIntegrating over $0<u<1$ proves the one-dimensional Prékopa–Leindler inequality.

For higher dimensions, use mathematical induction and Tonelli theorem. Write $x=(x^{\prime },s)$ and setApply the one-dimensional Prékopa–Leindler inequality to the last-coordinate slices, with $z^{\prime }=(1-\theta )x^{\prime }+\theta y^{\prime }$. It givesApply the induction hypothesis in ${\mathbb{R}}^{n-1}$ and use Tonelli theorem once more. For finite total integrals, infinite slice integrals occur only on null sets and may be replaced by zero there; this only weakens the displayed premise. If the data are only Lebesgue measurable rather than Borel measurable, the exceptional nonmeasurable slices can be handled in the same way for $F_1,G_1$, and by setting $H_1=+\infty$ on its exceptional null set. The total integrals are unchanged, and the slice premise holds everywhere it is needed. The result therefore holds for general nonnegative measurable functions.

The arithmetic-geometric mean inequality gives $a_k+1/a_k\ge 2$, so $u_n\ge 2n\to +\infty$. Hence $(u_n)$ has no finite limit.

For the example, take $\boxed{a_k=1+2^{-k}}$. Every term is positive and different from $1$, andThe bounding geometric series converges, so the comparison test for series proves that the partial sums $v_n$ converge.

In general, if $v_n$ converges, the necessary condition for a convergent series gives $d_k=a_k-1/a_k=v_k-v_{k-1}\to 0$. Since $a_k>0$,Apply the Cesaro theorem for convergent sequences to these summands:Equivalently, $a_k=(d_k+\sqrt{d_k^2+4})/2\to 1$. The series of the differences may converge while the series of the positive sums still diverges linearly.

Write $K=k_1+k_2+k_3>0$ and $C=ϵ+1-c_s^2$. The inflationary scalar Fourier mode has real late-time value $A_k=H/\sqrt{4ϵM_{\mathrm{Pl}}^2{c}_s{k}^3}$ and $u_k^{\prime \ast }(\tau )=A_k{c}_s^2{k}^2\tau e^{ik{c}_s\tau }$. Thus the time integrand, before the factor $-2i$, isThe Bunch-Davies vacuum prescription supplies damping in the past. Equivalently, use a positive regulator $d$ and then take the limit:The real part of $-2i$ times this integral is $4/(c_s^3{K}^3)$. Including the six Wick contractions yields the cubic time-derivative curvature bispectrumThe resulting primordial bispectrum is a homogeneous function of degree $-6$ in the wavenumbers, as expected for scale-independent background parameters.

The printed limit $c_s\ll 0$ is not a physical sound-speed limit for these modes. The normalization and positive-frequency choice assume $c_s>0$; a negative value is outside the positive sound-speed convention used by these modes. The likely intended limit is $0<c_s\ll 1$, which is analyzed explicitly here rather than silently substituted for the PDF.

To compare primordial non-Gaussianity, normalize by the power spectrum, not only by the raw three-point amplitude. HereAt an equilateral bispectrum configuration, the exact normalized amplitude isThus $c_s\to 1$ gives an amplitude of order $ϵ$, consistent with weak non-Gaussianity in single-field slow-roll inflation. For $0<c_s\ll 1$, this vertex gives an amplitude of order one, enhanced relative to the slow-roll limit but not diverging as $c_s^{-2}$ after power-spectrum normalization. At fixed $H,ϵ,M_{\mathrm{Pl}}$ the raw bispectrum does scale as $c_s^{-2}$, but $P_{\zeta }^2$ has the same scaling. The often-quoted larger small-sound-speed amplitudes of more general inflationary actions require their actual interaction coefficients and other vertices; they do not follow from the lone supplied vertex. These conclusions concern its tree contribution within the stated weak-perturbation approximation.

The bispectrum shape function isAt fixed perimeter, the arithmetic-geometric mean inequality bounds this by $1/27$, attained only at an equilateral bispectrum configuration. In a squeezed bispectrum configuration with $k_1=q\ll k_2\simeq k_3=k$, it behaves as $q/(8k)\to 0$. At a flattened bispectrum configuration $(k,k,2k)$ it is finite, $1/32$, with no boundary singularity. Thus the shape prefers equilateral triangles and suppresses squeezed triangles, with a broad finite flattened contribution. This is a shape preference, not a complete signal-to-noise calculation: actual observational weighting also depends on cosmological transfer functions, covariance and accessible triangles. If $C=0$, this vertex produces no bispectrum at any configuration; if $C<0$, the signed shape reverses while its absolute triangle preference is unchanged.