# Tag Info

14

$(f(x))^2 = 4x^2\left(1+\sqrt{1-x^2}\right)^2$. Let $y = \sqrt{1-x^2} \Rightarrow x^2 = 1 - y^2 \Rightarrow (f(x))^2 = 4(1-y^2)(1+y)^2 = 4(1-y)(1+y)^3$. Apply AM-GM inequality we have: 2 = (1-y) + \dfrac{1+y}{3} + \dfrac{1+y}{3} + \dfrac{1+y}{3} \geq 4\sqrt[4]{\dfrac{(1-y)(1+y)^3}{27}} \Rightarrow \dfrac{16\times 27}{4^4} \geq (1-y)(1+y)^3 \Rightarrow ... 11 We have, \begin{align}\int_0^1|f'(x)-f(x)|dx &= \int_0^1|f'(x)e^{-x}-f(x)e^{-x}|e^xdx \\&\geq \int_0^1\left(f'(x)e^{-x}-f(x)e^{-x}\right) dx\\&= \int_0^1 \frac{d\left(f(x)e^{-x}\right)}{dx}dx\\&= f(1)e^{-1}-f(0)e^{0}\\&=\frac{1}{e}\end{align} 9 I'll admit its essentially same as what ah-huh-moment did here, but here's another approach: Make the substitutionx = \sin \theta$, for$\displaystyle \theta \in \left(0,\pi/2\right)$, then,$\displaystyle f(x) = x + 2x\sqrt{1-x^2} = 2\sin \theta + \sin (\pi - 2\theta) \le 3\sin \frac{2\theta + \pi - 2\theta}{3} = \frac{3\sqrt{3}}{2}$from Jensen's ... 8 Observe that as long as$a,b,c,d \ge 0: \begin{align}(1-a)(1-b) - cd &\ge 1 - a - b + ab - \left(\frac{c^2+d^2}{2}\right) \\ &= 1 - a - b + ab - \left(\frac{1-a^2-b^2}{2}\right) \\ &= \frac{1}{2}(1-a-b)^2 \ge 0\end{align} Similarly, $$(1-c)(1-d) -ab \ge \frac{1}{2}(1-c-d)^2\ge 0$$ Hence, $$(1-a)(1-b)(1-c)(1-d) \ge abcd$$ 7 Use the fact that\log $is monotone increasing function: $$(a-b)(\log(a)-\log(b))\ge0 \implies \left(\frac{a}{b}\right)^{a-b} \ge 1$$ $$(c-a)(\log(c)-\log(a))\ge0 \implies \left(\frac{b}{c}\right)^{b-c} \ge 1$$ $$(b-c)(\log(b)-\log(c))\ge0 \implies \left(\frac{c}{a}\right)^{c-a} \ge 1$$ Multiply the three inequality yields your desired form ! 5 Apply Cauchy-Schwarz Inequality:$\left(4x+3y+2z\right)^2 = \left(2\sqrt{2}\cdot \sqrt{2}x+\sqrt{3}\cdot \sqrt{3}y+1\cdot 2z\right)^2 \leq \left((2\sqrt{2})^2+(\sqrt{3})^2+1^2\right)\cdot\left(2x^2+3y^2+4z^2\right) = 12 \Rightarrow 4x+3y+2z \leq 2\sqrt{3} = \text{Maximum Value}$. 5 If$D$is a diagonal matrix with$(b_1,\ldots,b_n)$in the diagonal and$b_i>0$then$tr(AD)=a_{11}b_1+\ldots a_{nn}b_n$. Since$a_{ii}>0$and$b_i>0$then$a_{11}b_1+\ldots +a_{nn}b_n\leq(a_{11}+\ldots +a_{nn})(b_1+\ldots+b_n)=tr(A)tr(D)$. If$B$is symmetric then$B=RDR^t$, where$R$is orthogonal and$D$is diagonal, and ... 5 This inequality is equivalent to$a\ln a+b\ln b\geq a\ln b+b\ln a$, which is obvious by Rearrangement or it's$(a-b)(\ln a-\ln b)\geq$, which is a proof of Rearrangement. 4 As Najib Idrissi pointed out in chat, the argument given by the answers in this question solves your question as well ! Substitute$x = \frac{1}{2}$in the final inequality. Integrate by parts to verify that for$f(0) = f(1) = 0$, $$f'(x) = \int_{0}^{x}tf''(t)\,dt - \int_{1}^{x}(t-1)f''(t)\,dt$$ Take absolute value on both sides and use triangle ... 4 By the triangle inequality,$|a + b| \le |a| + |b| \le |a| + 3|b|$. Thus $$|a + b| - |a| - |b| \le 2|b|$$ On the other hand,$|a + b| \ge |a| - |b|$, so $$|a + b| - |a| - |b| \ge 0 \ge -2|b|$$ Therefore $$|a+b| - |a| - |b| \ge -2|b|$$ Consequently,$||a + b| - |a| - |b|| \le 2|b|$. 3 Hint:$(2a-1)(2b-1)=4ab-2(a+b)+1$. 3 For such$f$there holds an estimate$f'(x)^2\le 2Mf(x)$for all$x\in \mathbb R$, where$M=\max|f''|$. See lemma 5.6 in Yu. V. Egorov, Linear Differential Equations of Principal Type. The idea is to consider$f$on an interval$(x_1,x_2)$where$x_1$and$x_2$are consecutive zeros of$f$. If the maximum of$f'^2/f$on$[x_1,x_2]$is attained at some ... 3 We may exploit an integral representation for$\gamma$and some trivial bounds. Since: $$\begin{eqnarray*}\gamma&=&\lim_{n\to +\infty}(H_n-\log n) = \lim_{n\to +\infty}\left(\int_{0}^{1}\frac{x^n-1}{x-1}\,dx-\int_{1}^{n}\frac{dt}{t}\right)\\&=&\lim_{n\to ... 3 We need to prove that$$\frac{k+1}{k}<\sqrt 2\implies \frac{k+2}{k+1}<\sqrt 2,$$Equivalent to saying that$$1+\frac 1k < \sqrt 2\implies 1+\frac 1{k+1}<\sqrt 2.$$The last one is obvious as \dfrac 1k>\dfrac 1{k+1}. 3$$x^3+y^3 - 4(x-y) \ge 2(x^2 - xy +y^2) - 4(x-y) = 2(x^2 - 2xy +y^2 - 2(x-y) + xy) \ge 2(x-y-1)^2$$Where, we used x+y \ge 2\sqrt{xy} \ge 2 and xy \ge 1 2 note This proof I started aimed to use induction, but turned into a direct proof as I never used the induction hypothesis. Base case: n=3, we have (3+1)^4 = 256 < 324 = 4\cdot 81 = 4\cdot (3)^4 Assume it is true for our induction hypothesis that (n+1)^4 < 4n^4 for some n\geq 3 We continue to try to show it is true for n+1 as well. ... 2 It's enough to prove that |1 + R^2e^{2i \theta}| \geq R^2 - 1. This comes easily from the reverse triangle inequality:$$ |1 + R^2 e^{2i\theta}| \geq |R^2 e^{2i\theta}| - |1| = R^2 - 1.$$2 One cute application is to the well-known lemma: Lemma: Let \{x_n\} be a sequence of positive real numbers. Then$$\prod_{n=1}^{\infty} (1+x_n)$$converges if and only if$$\sum_{n=1}^{\infty} x_n$$converges. Proof: One direction is clear, as \prod_{n=1}^{N} (1+x_n) > \sum_{n=1}^{N} x_n. For the other direction, use the AM-GM inequality. ... 2 First, change the ellipsoid to a sphere by writing v = x \sqrt{2}, w=y\sqrt{3}, u = 2z THe function to be maximized will still be linear, now in u, v, w.$$ a_u u + a_v v + a_w w $$Now proceed in the direction perpendicular to that plane, and lets look at a lihne thru the origin in that direction, which could be parameterized as$$ u = s/a_u \\ v = ... 2 For me,$4x + 3y + 2z = \langle\begin{pmatrix} \sqrt{2} x \\ \sqrt{3}y \\ 2z\end{pmatrix}, \begin{pmatrix} 2\sqrt{2} \\\sqrt{3} \\\ 1\end{pmatrix}\rangle \leq 1\times 2\sqrt{3}=2\sqrt{3}$2 Just another way: sum the AM-GMs... $$2x^2+2k^2 \ge k\cdot 4 x,\quad 3y^2+\frac34k^2 \ge k \cdot 3 y, \quad 4z^2+ \frac14k^2 \ge k\cdot2z$$ to get$\dfrac1k+3k \ge (4x+3y+2z)$, and the maximum is achieved when$\frac1k=3k\implies k = \frac1{\sqrt3}$, for a maximum of$2\sqrt3$. 2 A general linear algebra result says that for$S\succeq 0$,$R\succeq 0$such that$R-S\succeq 0$, we may infer that$|S|\leq |R|$. Apply this with $$S\equiv C^TC-C^TB(B^TB)^{-1}B^TC\succeq 0,\quad R\equiv C^TC\succeq0,\quad R-S\succeq0.$$ 2 For$p > 1$, apply Holder's Inequality to $$\left\lvert \sum_{k=1}^n x_i\right\rvert \leq \sum_{k=1}^n 1\cdot \lvert x_i\rvert \leq n^{1/q} \left(\sum_{k=1}^n \lvert x_i\rvert^p\right)^{1/p}$$ with$q =\frac{p}{p-1}$is the Holder conjugate of$p$. Now, raise both sides to the power$p$to get the inequality. For$p \in (0,1]$, this comes from the ... 2 Here's a different argument. Define $$X_t = \int_0^t f(s,\omega) dW_s \\ Y_t = X_t^4 \\ Z_t = X_t^2.$$ Apply the Ito formula to$Y_t$and$Z_t$: $$Y_t = 4 \int_0^t f(s,\omega) X_s^3 dW_s + 6 \int_0^t f(s,\omega)^2 X_s^2 ds \\ Z_t = 2 \int_0^t f(s,\omega) X_s dW_s + \int_0^t f(s,\omega)^2 ds$$ Now substitute and take expectations, thereby canceling the ... 2 By AM-GM inequality: $$\frac{ab}{a^2+b^2}= \frac12-\frac{(a-b)^2}{2(a^2+b^2)}\ge \frac12-\frac{(a-b)^2}{4ab}=1-\frac14\left(\frac{a}b+\frac{b}a\right)$$ Also $$\frac1{4a}= \frac{a+b+c}{4a}=\frac14+\frac14\left(\frac{b}a+\frac{c}a \right)$$ Cyclically summing the above two and adding, we get the required inequality. 2 For simplicity, lets say that we have two constants,$a,b: a\geq 1, b\leq 1$and that$P(Y=a)=p, P(Y=b)=(1-p)=q$with the constraint that$ap+bq=1=(a-b)p+b \implies p=\frac{1-b}{a-b}$Now,$\lim_{n\rightarrow \infty}X_n=0\; a.s. \iff P(\lim_{n\rightarrow \infty}X_n>0)=0$. As you pointed out, that is equivalent to requiring$\lim_{n\rightarrow ...

1

They made the substitution $(a_n + c_n )^2 + (b_n + d_n )^2 = -2a_nb_n - 2b_nc_n - 2c_nd_n - 2d_na_n$ in the equality: $$(a_{n+1}^2 + b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2) \\= 2(a_n^2 + b_n^2 + c_n^2 + d_n^2 ) - 2a_nb_n - 2b_nc_n - 2c_nd_n - 2d_na_n$$ to write it as: $$a_{n+1}^2 + b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 = 2(a_n^2 + b_n^2 + c_n^2 + d_n^2 ) + (a_n ... 1 As you obtained,$$0 < ( \lambda y_1- x_1)^2 + ( \lambda y_2- x_2)^2 \iff (y_1^2+y_2)^2\lambda^2-2(x_1y_1+x_2y_2)\lambda+(x_1^2+x_2^2) > 0$$Now if you are not familiar with using the discriminant, you can "complete the square" to get:$$\iff (y_1^2+y_2^2)\left(\color{red}{\lambda - \frac{x_1y_1+x_2y_2}{y_1^2+y_2^2}} ...

1

One of the proofs is the following that uses quadratic function which I like: $0\leq \displaystyle \sum_{j=1}^n(a_{1j}-\lambda x_j)^2 = \displaystyle \sum_{j=1}^n a_{1j}^2 - 2\lambda\displaystyle \sum_{j=1}^n a_{1j}x_j + \lambda^2\displaystyle \sum_{j=1}^n x_j^2 = f(\lambda), \forall \lambda \in \mathbb{R} \Rightarrow \triangle' \leq 0 \Rightarrow ... 1 We'll show a bit more:$a^a b^b > \left( \frac{a+b}{2} \right)^{a+b} > a^b b^a.$First we need a lemma:$P = (1+x)^{1+x} (1-x)^{1-x} > 1$if$x < 1$. Proof:$\ln P = (1+x)\ln (1+x) + (1-x) \ln (1-x) = x ( \ln (1+x) - \ln (1-x)) + \ln (1+x) + \ln (1-x) = 2x \left(x + \frac{x^3}{3}+\frac{x^5}{5} + \cdots \right) - 2 \left( \frac{x^2}{2} + ...

Only top voted, non community-wiki answers of a minimum length are eligible