# Tag Info

10

We shall show that $\,\Big(1+\sqrt{\frac{2}{n}}\,\Big)^n>n$. For $n=1$ it is obvious. Assume that $n\ge 2$, then according to the binomial theorem: \begin{align} \left(1+\sqrt{\frac{2}{n}}\right)^n&=\binom{n}{0}+\binom{n}{1}\sqrt{\frac{2}{n}}+\binom{n}{2}\left(\sqrt{\frac{2}{n}}\right)^{\!2}+\cdots+\binom{n}{n}\left(\sqrt{\frac{2}{n}}\right)^{\!n} ...

6

I think I'm found nice solving method. If $x\neq 0$: $$\frac{2x}{x^2-2x+5} + \frac{3x}{x^2+2x+5} \leq \frac{7}{8} \equiv \frac{2}{x-2+\frac{5}{x}} + \frac{3}{x+2+\frac{5}{x}} \leq \frac{7}{8}$$ Let $t=x+\frac{5}{x}$. $$\frac{2}{t-2} + \frac{3}{t+2} \leq \frac{7}{8}$$ $$\frac{(t-6)(7t+2)}{(t-2)(t+2)} \geq 0$$ $t \in (-\infty; -2) \cup [-\frac{2}{7}; 2) ... 5 Without loss of generality, let$a+b+c=3$. Then we only have to prove $$(a^2b+b^2c+c^2a)(ab+bc+ac)\le 9$$ Assume$b=\text{mid}(a,b,c)$, then we have $$c(b-a)(b-c)\le 0\Longleftrightarrow a^2b+b^2c+c^2a\le b(a^2+ac+c^2)$$ So $$(a^2b+b^2c+c^2a)(ab+bc+ac)\le b(a^2+ac+c^2)(ab+bc+ac)$$ Using AM-GM inequality, we get$\begin{align*} ...

5

Notice that $$3(a+b+c)abc(a^2+b^2+c^2) \leq (ab+bc+ca)^2(a^2+b^2+c^2) \leq \left(\frac{(a+b+c)^2}{3}\right)^3.$$ The first inequality follows from $a^2b^2+b^2c^2+c^2a^2 \geq abc(a+b+c)$, which is true as $a^2b^2+b^2c^2 \geq 2ab^2c$, $b^2c^2+c^2a^2 \geq 2abc^2$ and $c^2a^2+a^2b^2 \geq 2a^2bc$ by AM-GM. The second inequality follows from AM-GM applied on ...

5

since $$\sum_{cyc}\dfrac{a}{b+c^2}=\sum_{cyc}\dfrac{a^2}{ab+ac^2}$$ Use Cauchy-Schwarz inequality,we have $$\left(\sum_{cyc}\dfrac{a^2}{ab+ac^2}\right)\cdot\sum_{cyc}(ab+ac^2)\ge\left(\sum_{cyc}a\right)^2=1$$ so we only prove this following inequality $$\dfrac{1}{\sum_{cyc}(ab+ac^2)}\ge\dfrac{9}{4}$$ $$\Longleftrightarrow 4\ge9\sum_{cyc}ab+9\sum_{cyc}ac^2$$ ...

4

The inequality is equivalent to: $$\left(\frac{1+a_1}{2}\right)\cdots \left(\frac{1+a_n}{2}\right) \le \left(\frac{1+a_1\cdots a_n}{2}\right)$$ Assuming the induction hypothesis we want to show: $$\left(\frac{1+a_1}{2}\right)\cdots \left(\frac{1+a_{n+1}}{2}\right) \le \left(\frac{1+a_1\cdots a_{n+1}}{2}\right)$$ It is enough to show: $$... 4 The statement holds for any increasing f:[0,1]\to[0,+\infty), i.e. the continuity of f is not necessary. Note that if the statement holds for some f, then after adding a common positive constant to f,g,h simultaneously, the statement still holds, so for simplicity, we may assume f(0)=0. Following your notations, define H by letting H(x)= 1 ... 4 I think I got a very nice proof. Consider that:$$\tan(\sin x)=\int_{0}^{\sin x}\frac{d\theta}{\cos^2\theta}=\int_{0}^{\sin x}\frac{d\theta}{1-\sin^2\theta}=\int_{0}^{x}\frac{dy}{(1-y^2)^{3/2}},$$while:$$\sin(\tan x)=\int_{0}^{\tan x}\cos\theta\,d\theta = \int_{0}^{x}\frac{dy}{(1+y^2)^{3/2}},$$so the inequality \tan(\sin x)> \sin(\tan x) is trivial ... 3 Considering the problem from an algebraic point of view, the equation$$ x/ 2^x=2 $$has no simple solution except in terms of Lambert function. The explicit solution is given by$$x=-\frac{W(-2 \log (2))}{\log (2)}$$which is a complex number (x=0.12792 -2.18169 i). So, even if \mathbb R, there is no solution (as mentioned by dani_s). 3 Let t = \frac xy > 1. Then:$$\begin{align} \frac{\log x}x = \frac{\log t + \log y}{ty} < \frac{\log y}y &\iff \log t + \log y < t\log y\\ &\iff \log t < (t - 1)\log y\\ &\iff \frac{\log t}{t-1} < \log y\\ &\iff\frac{\log [(t-1) + 1]}{t-1} < \log y. \end{align}$$Put z = t-1 > 0, then we prove: \frac{\log(z+1)}z ... 2 Let x = b(1)/a(1) > 0 and y = b(2)/a(2) > 0. Then we need to prove: ((x*a(1) + y*a(2))/(a(1) + a(2)))^(a(1) + a(2)) > x^(a(1))*y^(a(2) <====> (a(1) + a(2))*ln(x*a(1)/(a(1) + a(2)) + y*a(2)/(a(1) + a(2))) > (a(1)/(a(1) + a(2)))*lnx + (a(2)/(a(1) + a(2)))*lny. Let m = a(1)/(a(1) + a(2)) > 0, and n = a(2)/(a(1) + a(2)) > 0 then m + n = 1. So we ... 2 Claim.$$ S_N(x)=\sum_{k=1}^N \frac{\sin (kx)}{k} > 0 $$I just give the main ideas, I hope you will succeed with it. Arguing by contradiction, consider a point x_0 \in (0,\pi) where the sum S_N(x)=\sum_{k=1}^N \frac{\sin (kx)}{k} reaches a negative minimum Using the necessary condition for the existence of a minimum (S_n'(x_0)=0) show ... 2 As others have pointed out, you should prove this by using the triangle inequality. I also think you should try to understand the problem intuitively so I drew a picture: |a-b| represents the distance between the points a and b on the number line. If c is between a and b, or is equal to either a or b, then the distance from a to c, which ... 2 I followed your induction strategy. Starting from:$$(1+a_1)\dots(1+a_{k+1})\le 2^{k-1}(a_1\dots a_k+1)(a_{k+1}+1)$$Let's take A = a_1 \dots a_k.$$2^{k-1}(a_1\dots a_k+1)(a_{k+1}+1)=2^{k}(A.a_{k+1} + 1)+2^{k-1}(A-A.a_{k+1}-1+a_{k+1})$$The first part of the right formula is what we want. We have to prove that the second part is negative to prove ... 2 Answer: Multiplying a in the numerator and the denominator, you get$$LHS = \frac{\sum a^{2}}{\sum (ac+b^{2}a)}$$Applying Cauchy Schwarz inequality in the below steps$$>=\frac{(\sum a)^{2}}{\sum ac + \sum b^{2}a}>=\frac{1}{\sum ac + \frac{1}{3}\sum a \sum a^2} = \frac{1}{3\sum ac +\sum a (\sum a)^2}>=\frac{3}{\sum ac + (\sum ...

2

This post is due to the reason that no-one elaborated the AM-GM technique, and any beginner not knowing this method might get help to learn this method of proof. $\dfrac{x+\frac{1}{x}}{2}\ge \sqrt{x\cdot \dfrac{1}{x}} \implies x+\dfrac{1}{x}\ge 2$ Please note that this is a direct consequence of- a perfect square is always postive.

2

Since $x\gt 0$ you can multiply through by $x$ to clear fractions without changing the sense of the inequality. This gives $$x^2+1\ge2x$$Subtract $2x$ from each side:$$x^2-2x+1\ge0$$ or $$(x-1)^2\ge0$$ Which is true, with equality only if $x=1$ since squares are non-negative. Now note that each of these steps can be reversed to take us from the last ...

2

The inequality can be written as $$(1+d)\left(\sum_{\text{cyc}}\frac{a}{\sqrt[3]{b^3+abcd}}\right)^3\geq27$$ By the generalized Hölder's inequality $$\left(\sum_{\text{cyc}}\frac{a}{\sqrt[3]{b(b^2+d\,ca)}}\right)^3\left(\sum_{\text{cyc}}ab\right)\left(\sum_{\text{cyc}}a(b^2+d\,ca)\right)\geq\left(\sum_{\text{cyc}}a\right)^5$$ This reduces the problem to ...

1

We have that $$\frac{a_n}{b_n}\ge \frac{a_i}{b_i}\iff a_nb_i\ge a_ib_n\;\;,\;\;\forall i\implies$$ $$\frac{a_n}{b_n}\ge\frac{a_1+\ldots+a_n}{b_1+\ldots+b_n}\iff a_nb_1+\ldots +a_nb_n\ge a_1b_n+\ldots+a_nb_n$$ and the claim follows from the fist part above Now you try the other inequality.

1

No, actually the opposite inequality should be true, that is $$λ\le λ_L$$ To see that, fix $x\in \mathbb R$. Then minimization of the function $f(x,\cdot)$ with respect to $y$ over a larger set, i.e. $$\{y\in \mathbb R: g(y) \in \mathbb R\}$$ (which of course includes the case $g(y)=c$), yields a minimum that is less or (at most) equal to the minimum of the ...

1

There're no solutions $n\in\Bbb R$. Clearly, $n$ can't be negative. In zero we have $$n<2^{1+n}.$$ Left hand side has a derivative with respect to $n$ equal to $1$. Right hand side has a derivative with respect to $n$ equal to $2^{1+n}\ln 2$ which is obviously greater than $1$ for all $n\ge 0$. Therefore, $\forall n\in \Bbb R$ we have ...

1

Another (easy) way to see that there are no real solutions to this problem is plotting the two functions \left\{\begin{aligned} y&=n\\ y&=2^{n+1} \end{aligned}\right. which will eventually meet where your equation has solutions. You can do it in many ways. Probably the most easy way to do that is using WolframAlpha and asking literally for it ...

1

The reasoning of the solution is the following. Firstly, the inequalities $(\star)$ and $(\dagger)$ represent the two principle minors of the Hessian matrix of $f$. It is known that $f$ is convex iff it's Hessian matrix is positive semidefinite and the Hessian matrix is positive semidefinite if it's principle minors are positive. So, that explains the ...

1

Rewriting this in terms of $X\setminus A_j$ and using $\mu(X)=1$, the inequality is equivalent to $$\mu\left(\bigcup_{j=1}^n(X\setminus A_j)\right)\leqslant \sum_{j=1}^n\mu(X\setminus A_j).$$ This can be handled integrating the inequality $$\chi\left(\bigcup_{j=1}^nB_j\right)\leqslant \sum_{j=1}^n\chi(B_j)$$ valid for any collection $(B_j)_{j=1}^n$ of ...

1

Another approach would be to consider a factored form of the inequality: $(x-4)(x+4)>0$ Now, for this to be positive, both factors have to share the same sign and so one could look at this as 2 cases: 1) $x-4>0$ and $x+4>0$ which would imply $x>4$ 2) $x-4<0$ and $x+4<0$ which would imply $x<-4$ Then, one just combines these cases to ...

1

Squaring is the best method to prove the inequality. However, there is a detour, which involves squaring and nastier computation (and somewhat magical factorization), that may be what you're looking for. Substitute $x = \frac{29}{\sqrt 2}$ in $t(x) = x^2 - 41x + 420$ to get \begin{align} t\left(\frac{29}{\sqrt 2}\right) = \frac{29^2}{2} - 41 \cdot ...

1

$$20<29/\sqrt{2}<21\\ 20<29/\sqrt{2}\mbox{ and }29/\sqrt{2}<21$$ Write $$29/\sqrt{2}=(29\cdot\sqrt{2})/2$$ Then, $$20<(29\cdot\sqrt{2})/2\mbox{ and }(29\cdot\sqrt{2})/2<21\\ 40<(29\cdot\sqrt{2})\mbox{ and }(29\cdot\sqrt{2})<42\\ 40/29<\sqrt{2}\mbox{ and }\sqrt{2}<42/29\\ \mbox{Thus }40/29<\sqrt{2}<42/29$$ Q.E.D.

1

$$f(x) = \frac{x}{\lfloor x\rfloor}$$ Solve $f(x) \ge \frac32$. Since $f(x) \gt 0$(why?) we can write, $$\frac{1}{f(x)} = \frac{\lfloor x\rfloor}{x} \le\frac23 \implies \frac{x-\{x\}}{x}\le\frac23$$, where $\{x\}$ is the fractional part of $x$ $$1-\frac{\{x\}}{x} \le \frac{2}{3} \implies \frac{\{x\}}{x} \ge \frac13$$ $$\{x\} \ge \frac{x}{3}$$ But since, ...

1

Since $\lfloor x\rfloor\le x$, if $x\lt0$, $$x\le\frac32\lfloor x\rfloor\le\frac32x$$ which simplifies to $x\ge0$. Therefore, $x\ge0$. Since $\lfloor x\rfloor\gt x-1$, $$x\ge\frac32\lfloor x\rfloor\gt\frac32(x-1)$$ which simplifies to $0\le x\lt3$. If $2\le x\lt3$, then $$\frac x2\ge\frac32\implies x\ge3$$ If $1\le x\lt2$, then  \frac ...

1

The answer is no. A pretty nice counter-example has been given by Stephen in this question: Friedrichs's inequality? Backstory 1: $\mathbf{H}_0(\operatorname{div};\Omega) \cap \mathbf{H}(\mathbf{curl};\Omega)$ and $\mathbf{H}(\operatorname{div};\Omega) \cap \mathbf{H}_{0}(\mathbf{curl};\Omega)$ are compactly embedded in $\mathbf{L}^2(\Omega)$. ...

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