Hot answers tagged

12

Both sums are norms, $\lVert. \rVert_2$ vs $\lVert . \rVert_1$ and thanks to norm equivalence for finite dimensional spaces there are factors which fulfill $$ m \lVert a \rVert_2 \le \lVert a \rVert_1 \le M \lVert a \rVert_2 $$ So having $\lVert a \rVert_2^2 = x$ we get $$ m \sqrt{x} \le \lVert a \rVert_1 \le M \sqrt{x} $$ for $m =1$ and $M = \sqrt{n}$.


7

You can use the AQM inequality: $$\frac{\displaystyle\sum\nolimits_{i=1}^n\lvert a_i\rvert}n \le \sqrt{\frac{\displaystyle\sum\nolimits_{i=1}^n a_i^2}n}=\sqrt{\frac xn}, \enspace\text{whence}\enspace \displaystyle\sum_{i=1}^n\lvert a_i\rvert\le\sqrt{nx\mathstrut}.$$


7

First note that the series converges using Leibniz Test. Next, denote by $S_N$ the partial sum $\sum_{n=1}^N\frac{(-1)^{n-1}}{n}$. Then, we must have $$S_{2N}<\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}<S_{2N+1}$$ Finally, we see that $\sum_{n=1}^4 \frac{(-1)^{n+1}}{n}=\frac{7}{12}$ and inasmuch as the next term is positive, the value of the series must ...


6

Happily for you, it turns out that $\sqrt{\sum_{i=1}^n a_i^2 } \leq \sum_{i=1}^n |a_i| \leq \sqrt{n} \cdot \sqrt{\sum_{i=1}^n a_i^2 }$ For more information on this, see https://en.wikipedia.org/wiki/Norm_(mathematics)


5

Here is an argument which borrowed ideas from measure theory, but does not assume any direct knowledge on it. Proof. It suffices to prove the following claim: Claim. If $h \in \mathscr{R}([a, b])$ satisfies $h \geq 0$ and $\int_{a}^{b} h \, \mathrm{d}x = 0$, then $h(c) = 0$ for some $c \in [a, b]$. Step 1. To this end, assume that $h$ satisfies the ...


5

Without any special treatment for $x=\frac{\pi}{2}$, which can be verified manually, we have $\forall x \in \left [0, \frac{\pi}{2} \right )$: $$0<\cos(x)\leq e^{-\frac{x^2}{2}} \Leftrightarrow \ln(\cos(x)) \leq -\frac{x^2}{2}$$ just because $\ln(x)$ is ascending, strictly. As a result, let's look at this function $f(x)=-\frac{x^2}{2}-\ln(\cos(x))$. ...


4

All you need to do is a little algebra to substitute $f_1 + f_2$ into the new definition. To apply Minkowski inequality, let's denote the traditional $L_2$ norm of $f$: $\left(\int_0^1 |f|^2 dx\right)^{1/2}$ by $\|f\|_0$. We have: \begin{align} & \|f_1 + f_2\|^2 \\ = & \int_0^1 |f_1 + f_2|^2 dx + \int_0^1 |f_1' + f_2'|^2 dx \\ = & \|f_1 + ...


4

For any positive real numbers $A_1,A_2,B_1,B_2$ AM-GM inequality implies \begin{align*} 2\sqrt{A_1A_2B_1B_2}&\le A_1B_2+A_2B_1 \end{align*} Then \begin{align*} A_1A_2+2\sqrt{A_1A_2B_1B_2}+B_1B_2&\le A_1A_2+A_1B_2+A_2B_1+B_1B_2\\ \left(\sqrt{A_1A_2}+\sqrt{B_1B_2}\right)^2&\le\left(A_1+B_1\right)\left(A_2+B_2\right) \end{align*} So ...


4

If you have proved that $\beta <-1$ then you have proved that $|\beta| \gt 1$. Since $\alpha\beta=1$, $|\alpha| |\beta|=1$. So if $|\beta| \gt 1$ then $|\alpha| \lt 1$


3

If $s_{n}:=\sum_{k=1}^{n}\frac{\left(-1\right)^{k+1}}{k}$ then $\left(s_{2n-1}\right)_{n=1,2,\dots}$ is monotonically decreasing and $\left(s_{2n}\right)_{n=1,2,\dots}$ is monotonically increasing. This with $s_{2n}<s_{2n-1}$. So if you can find some $n$ with $\frac{7}{12}<s_{2n}<s_{2n-1}<\frac{47}{60}$ then you are ready.


3

You may use the Weierstrass product for the cosine function: $$ \cos x = \prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right) \tag{1}$$ and by taking $\log$s: $$\begin{eqnarray*}\log\cos x&=&\sum_{n\geq 0}\log\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right)\\&=&-\sum_{n\geq 0}\sum_{m\geq 1}\frac{4^m ...


3

By Heron's formula $$ 16\Delta^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c) \tag{1}$$ and by the AM-GM inequality $$ (-a+b+c)(a-b+c)(a+b-c) \leq \left(\frac{a+b+c}{3}\right)^3 \tag{2} $$ Equality is attained only at $a=b=c$.


3

Use Cauchy-Schwarz inequality: $$\left(a+b\sqrt{2}\sin{x}+c\sin{2x}\right)^2\le (a^2+b^2+c^2)(1+2\sin^2x+\sin^22x)$$ $$\left(a+b\sqrt{2}\sin{x}+c\sin{2x}\right)^2\le 100\cdot(1+2\sin^2x+\sin^22x)$$ $$|a+b\sqrt{2}\sin{x}+c\sin{2x}|\le 10\cdot\sqrt{1+2\sin^2x+\sin^22x}$$ $$1\le1+2\sin^2x+\sin^22x\le \frac{13}{4}$$ Then $$-5\sqrt{13}\le ...


3

No. It is only true (i.e. it only needs to be true) if $j$ is the index of the minimum among all the $|z_k|$'s. Example: $z_1 = 1$, $z_2 = 2$ and $j=2$.


3

Hint: For $a>0$, we have by AM-GM that $4ax^2+\frac{1}{x}\geq 3\sqrt[3]{a}$. The equality holds iff $x=\frac{1}{2\sqrt[3]{a}}$. For $a<0$, see copperhat's comment. For $a=0$, this is trivial.


3

It is routine execrise to show that $$ t\rightarrow e^{t} $$ is a convex function by checking second derivative. Now define the function (for fixed $x)$ $$ t\rightarrow p(1-t)\log(|f(x)|)+qt\log(|g(x)|)\rightarrow e^{p(1-t)\log(|f(x)|)+qt\log(|g(x)|)} $$ and it is the composition of a linear function with a convex function. Thus it must be convex as well. ...


3

We have $$ab+ac+ad+bc+bd+cd=\frac12\bigl((a+b+c+d)^2-(a^2+b^2+c^2+d^2)\bigr) $$ and for $a+b+c+d=4$ we have $a^2+b^2+c^2+d^2\ge4$ (why?) so that $$ab+ac+ad+bc+bd+cd\le\frac{16-4}{2}= 6 $$ and you indeed obtain $\le 8$ for the original inequality.


3

Let $a+b+c=t$. Then $11a-t\ge d\ge t$, so that $a\ge \frac{2t}{11}$. Similarly, $b\ge \frac{2t}{6}$ and $c\ge \frac{2t}{3}$. Summing these up, we get: $$t=a+b+c\ge\frac{2t}{11}+\frac{2t}{6}+\frac{2t}{3}>t$$ which is impossible.


2

Let $x=0$. Then $y^2+z^3=1 \Rightarrow z^3=1-y^2$ Seek to maximise $p=y^2z=y^2(1-y^2)^{\frac 13}$ WLOG let $u=y^2$ so that $p=u(1-u)^{\frac 13}$ $\frac {dp}{du}=(1-u)^{\frac 13}-u{\frac 13}(1-u)^{-\frac 23}$ $\frac {dp}{du}=\frac 13(3-4u)(1-u)^{-\frac 23}$ Maximum is at $u=\frac 34$ Thus maximum is $p=\frac 34(\frac 14)^{\frac 13}$ $p \le \frac ...


2

I don't think the best bound can be solved for analytically. using Lagrange multipliers , I tried to maximize $x^2y+y^2z+z^2x$ $\quad$ subject to the constraint $x+y^2+z^3=1$ maximize $g=x^2y+y^2z+z^2x+\lambda(x+y^2+z^3-1)$ $\frac {\partial g}{\partial x}=0:2xy+z^2+\lambda=0$ $\frac {\partial g}{\partial y}=0:x^2+2yz+2\lambda y=0$ $\frac {\partial ...


2

The method to be employed here is exactly the same as in a previous answer: Olympiad inequality $\sum_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$ Let $u = x$ , $v = y^2$ , $w = z^3$ , then $u,v,w \ge 0$ , $u+v+w=1$ and the inequality to be established: $$ x^2y+y^2z+z^2x < \frac12 \quad \Longrightarrow \quad f(u,v,w) = ...


2

Edit: this is only a partial answer. $$\sum_{i,j=1}^{n}a_{i}a_{j}f\left(\dfrac{|x_{i}-x_{j}|}{2}\right)\le 3\sum_{i,j=1}^{n}a_{i}a_{j}f(|x_{i}-x_{j}|)$$ $$\sum_{i,j=1,i\neq j}^{n}a_{i}a_{j}f\left(\dfrac{|x_{i}-x_{j}|}{2}\right)\le 3\sum_{i,j=1,i\neq j}^{n}a_{i}a_{j}f(|x_{i}-x_{j}|) + 2\sum_{i=1}^{n}a_{i}a_{i}$$ $$\sum_{i,j=1,i\neq ...


2

Am I correct? You are not correct. There are solutions for $a\ge 2$. It seems that you don't understand the following : $(ae^x)/(2e^x-1)\lt 1\iff$ "$2e^x-1\gt 0$ and $ae^x\lt 2e^x-1$" or "$2e^x-1\lt 0$ and $ae^x\gt 2e^x-1$". You got the right answer for $0\lt a\lt 2$ (though "and" has to be "or"), but the process looks strange especially in case 2 ...


2

[Edited after After Michael's comment] In the general case, the OP's claim is not verified. For example if $$ X = \begin{bmatrix} 29& 11& 19\\ 11& 10& 12\\ 19& 12& 19 \end{bmatrix} , \quad Y = \begin{bmatrix} 14& 16& 17\\ 16& 21& 22\\ 17& 22& 29 \end{bmatrix} , \quad w = \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix} ...


2

This is not possible For simplicity, just take $r=2$ and the nonzero singular values of $C$ are $1,1$. So that the F norm of C is $\sqrt2$. Now you may put the nonzero singular values of $A$ as $1/N, N$ and those of $B$ as $N, 1/N$ aligned accordingly Then $A,B$ have the same F-norm $\sqrt{N^2+1/N^2}$, but $N$ can be as big as you like


2

As you wrote, from the identity: $$x(xy+yz+zx)+y^2+z^3-x^2y-y^2z-z^2x=z(\frac{z}{2}-x)^2+(y^2+\frac{3}{4}z^3-y^2z)+xyz$$ It suffices to check $y^2+\frac{3}{4}z^3\ge y^2z$ given $yz\le 1$. We take $3$ cases: If $z\le 1$, then $y^2\ge y^2z$ already. If $z\ge 1$ and $y\le \frac{1}{4}$, then $\frac{3}{4}z^3\ge \frac{3}{4}z\ge \frac{1}{16}z\ge y^2z$. If $z\ge ...


2

Note: in this solution, $\sum$ denotes the cyclic sum, so that $\sum f(a, b, c, d)=f(a, b, c, d)+f(b, c, d, a)+f(c, d, a, b)+f(d, a, b, c)$. Lemma: $\sum c^4+2\sum abc^2+6\sum ab\le 36$. Proof: Expanding the inequality $\sum[2(a-b)^2+(ab-1)^2]\ge 0$ gives: $$6\sum ab\le 20+\sum a^2b^2$$. Expanding $\sum (ab-ac)^2\ge 0$ gives: $$2\sum abc^2\le \sum ...


2

If $x=y=1,\ z=0.1$ the determinant is $-0.3957...$ And for $x=y=1,z=0.5$ the determinant is $+0.1781...$ So since it is continuous in $z$ for fixed $x,y$ there is a value of $z$ between $0.1$ and $0.5$ for which $(x,y,z)=(1,1,z)$ makes the determinant zero. Maybe the ordering in one of the rows should be different, but this can only be checked from the ...


2

By computing $Q(-e_i)$ for all $i$ you can see that all of $Q$'s entries must be nonpositive. This means that two columns of $Q$ are orthogonal only if they are structurally orthogonal; since all columns must have at least one nonzero entry, this implies that all of the columns have exactly one nonzero entry. Therefore the only $Q$ satisifying your ...


2

If we make a change of variable $a=1/A^3$ and $x=Au/2$, then the problem becomes one of finding the largest value of $A$ for which the inequality $$u^2+{2\over u}\ge A$$ holds for all $u\gt0$. If you know calculus, then it's easy to see that $f(u)=u^2+{2\over u}$ is minimized at $u=1$, so that $f(1)=3$ is the largest that $A$ can be. If you don't know ...



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