New answers tagged

3

Rearrange: $$\frac{xb+(1-x)c}{a}=\frac{xc+(1-x)a}{b}$$ $$xb^2+(1-x)bc=xac+(1-x)a^2$$ $$x(b^2-ac)=(1-x)(a^2-bc)$$ Repeating this for another equality we obtain:- $\frac{x}{1-x}=\frac{b^2-ac}{c^2-ab}=\frac{a^2-bc}{b^2-ac}$ $(b^2-ac)^2=(c^2-ab)(a^2-bc)$ $a^3+b^3+c^3=3abc$ $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$ $a+b+c=0$ or $(a-b)^2+(b-c)^2+(c-a)^2=0$ $a+b+c=...


3

Yes. Your desired inequality is a consequence of the one you mentioned. In other words, it follows from the fact that the ranks of both $AB$ and $BA$ are non-negative integers at most equal to that minimum.


1

Hint: use the property: $ \frac{\displaystyle x}{\displaystyle y} = \frac{\displaystyle z}{\displaystyle t} = \frac{\displaystyle m}{\displaystyle n} = \frac{\displaystyle x+z+m}{\displaystyle y+t+n}$


3

Let $p = x+y+z, \ q = xy+yz+zx, \ r = xyz$. Rewrite the inequality as $p^2 q^2 \le 3(p^2q^2 - q^3 - p^3 r)$. Since $q^2 \ge 3pr$, it suffices to prove that $$p^2 q^2 \le 3(p^2q^2 - q^3 - p^3 \frac{q^2}{3p})$$ or $$q^2(p^2-3q)\ge 0.$$ It is obvious. We are done.


0

Does not exist. Try $b=a$, $c=\frac{1}{a^2}$ and $a\rightarrow0^+$.


0

You arrive at that answer by (incorrectly) flipping a sign in the first inequality so it reads $$x^2\boxed{+}4x-77<0\text{.}$$ Then the solution to that inequality is $x\in(-11,7)$; mutatis mutandis you get $x=-10$.


1

$\quad3\left(y^2 + yz + z^2\right)\left(x^2 + xz + z^2\right)\left(x^2 + xy + y^2\right)-\left(x+y+z\right)^2\left(yz+xz+xy\right)^2\\=\sum_{sym} \left(2x^4y^2z^0+0.5x^3y^3z^0+0.5x^4y^1z^1-2x^3y^2z^1-x^2y^2z^2\right)\\=2\sum_{sym} \left(x^4y^2z^0-x^3y^2z^1\right)+0.5\sum_{sym} \left(x^3y^3z^0-x^2y^2z^2\right)+0.5\sum_{sym} \left(x^4y^1z^1-x^2y^2z^2\right)$ ...


2

You're done if you prove that $f'(x)>1$ for $x>2$. By setting $\pi/x=t$, this is the same as proving that $$ g(t)=\cos t+t\sin t>1,\qquad 0<t<\pi/2 $$ Note that $g'(t)=t\cos t>0$ and $\lim_{t\to0}g(t)=1$.


1

Another possible way of proving your inequality is rearranging and then applying the MVT as follows: \begin{eqnarray*} (x+1)\cos\left(\frac{\pi}{x+1}\right)-x\cos\left(\frac{\pi}{x}\right) & > & 1\\ & \Leftrightarrow & \\ x\left( \cos \frac{\pi}{x+1} - \cos \frac{\pi}{x} \right) & > & 1 - \cos \frac{\pi}{x+1} \\ & \stackrel{...


5

According to your work, $$f''(x)=-\frac{{\pi}^2}{x^3}\cos\left(\frac{\pi}{x}\right)<0$$ for all $x\in(2,+\infty)$. Hence $$f'(x)=\cos\left(\frac{\pi}{x}\right)+\frac{\pi}{x}\sin\left(\frac{\pi}{x}\right)$$ is strictly decreasing in $(2,+\infty)$. Since $\lim_{x\to +\infty}f'(x)=1$, it follows that $f'(t)>1$ for all $t> 2$. Finally, for $x\geq 2$, ...


1

We know there are at least $a$ positions before Petr and at most $b$ positions after Petr. Let $i$ be a possible position for Petr. Then $1,2,...,i-1$ are the positions in front of Petr and this number, $i-1$, must be at least $a$, i.e. $i-1\ge a\implies i\ge a+1$ which is equivalent to $a+1\le i$. The positions after Petr are $i+1, i+2, ..., n$ and the ...


0

Given $p+q=1$ Observe that $\frac{p+q}{2} \geq \sqrt{pq}$ Therefore :- $\frac1{4} \geq {pq}$ Also, $p^2+q^2 \geq 2pq$ similarily, $\frac1p^2+\frac1q^2 \geq \frac2{pq}$ Adding both the inequalities We can see $\big(p+\frac1p\big)^2+\big(q+\frac1q\big)^2 \geq \frac2{pq}+2{pq}+4$ You can see for getting minimum value the left hand side should be as minimum ...


0

First, I will ignore the constraint $a^ab^bc^c=1$ to sketch a possible solution for the aforementioned inequality, under an alternative constraint: Set $x=\frac{1}{a^2+b^2}$ and $x=\frac{1}{b^2+c^2}$ and $z=\frac{1}{c^2+a^2}$. From the aritmetic-geometric inequality $$ x^2+y^2+z^2\geq 3(x^2y^2z^2)^{\frac{1}{3}}, $$ it remains to show that $$(x^2y^2z^2)^{\...


-1

Let, $F$ be the finite field of size $2$ and let $K = F^n$. Of those $K$ subcodes of distance, $3$ fix a $K$ subcode $C$ of maximum dimension. The unit $K$ codewords together generate $K$ while the non-trivial quotient code $K/C$ has a Hamel basis $Z$ made of unit $K/C$ codewords; i.e. made of $C$ cosets that contain a unit $K$ codeword. $Z$ has at most ...


2

$$f'(x)=ax^{a-1}-\ln{(a)}a^x$$ we need to solve $$ax^{a-1}-\ln{(a)}a^x>0$$ which would define a $2D$ region where for a given value of $a$ would give the interval that satisfies above (the cross section at $a=c$ for constant $c$ in the graph) . Note for $a<0$ the solution is complex and for $a=0$ it is undefined. WolframAlpha gives you a free ...


1

Using the substitutions $(a, b, c) \to (a^2, b^2, c^2)$, the inequality becomes $$\sum_{\mathrm{cyc}}\frac{a^2}{\sqrt{a^4+3b^2c^2}} \le \frac{9(a^4+b^4+c^4)}{2(a^2+b^2+c^2)^2}.$$ Using the Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align} \mathrm{LHS}^2 &= \sum_{\mathrm{cyc}} \frac{a^4}{a^4 + 3b^2c^2} + \sum_{\mathrm{cyc}} \frac{2a^2b^2}{\sqrt{...


0

The suggested inequalities in the previous answer are incorrect if $\min \vec{x} > 1$. (Counter-example: $\vec{x} = \{ 2, 3, 5 \}$.) The corrected inequalities are \begin{align} \frac{m(r-1)}{n(r+1)} \leqslant AM - HM \leqslant m(\sqrt{r} - 1)^2, \end{align} where $m = \min \vec{x}$.


0

Let $y_k=\sqrt{1+\sum_{h=1}^kx_h^2},\cos(\theta_k)=\frac{x_k}{y_k}$, Since $y_k^2=y_{k-1}^2+x_k^2$, so $\sin(\theta_k)=\frac{y_{k-1}}{y_k}$ So $\frac{x_k}{1+\sum_{h=1}^kx_h^2}=\frac{\cos(\theta_k)}{y_k} = \frac{\sin(\theta_k)\cos(\theta_k)}{y_{k-1}}$ So LEFT = $\frac{\cos(\theta_1)}{y_1}+\frac{\cos(\theta_2)}{y_2}+...+\frac{\cos(\theta_n)}{y_n}$ $\frac{\...


1

Let us call the individual terms of the inequality as $A_i$, then by AM-RMS inequality we have $$\frac{\sum A_i}{n} \le \sqrt{\frac{\sum A_i^2}{n}},~\mbox{where} ~A_i=\frac{x_i}{1+x_1^2+x_2^2+x_3^2+...x_i^2}.~~~~(1)$$ So it would suffice to prove that $\sum A_i^2 \le 1.$ Note that for $i\ge 2$, $$ A_i^2=\frac{x_1^2}{(1+x_1^2+x_3^2+x_3^2+...x_i^2)^2}\le \...


2

Let $x_0=0$. Thus, by C-S $$\sum_{k=1}^n\frac{x_k}{1+\sum\limits_{i=1}^kx_i^2}\leq\sqrt{n\sum_{k=1}^n\frac{x_k^2}{\left(1+\sum\limits_{i=1}^kx_i^2\right)^2}}\leq$$ $$\leq\sqrt{n\sum_{k=1}^n\frac{x_k^2}{\left(1+\sum\limits_{i=0}^{k-1}x_i^2\right)\left(1+\sum\limits_{i=1}^kx_i^2\right)}}=\sqrt{n\left(\frac{\sum\limits_{k=1}^nx_k^2}{1+\sum\limits_{k=1}^nx_k^2}\...


0

If $h$ is not equal to zero almost everywhere, then $\alpha(h)=1$. First let us restrict ourselves to $f,g \in L^2(\mathbb{R})$. Otherwise we run into troubles with multiplying zero and infinity. I am assuming that we are talking about the Lebesgue measure with the usual Lebesgue sigma-algebra. Case 1: h equal to zero almost everywhere. Let's take care of ...


2

Hint: apply A.M >G.M between for (1-a)(1-b)(1-c) $\sqrt[3]{(1-a)(1-b)(1-c)} \le \cfrac{(1-a)+(1-b)+(1-c)}{3}$$


0

I found a way to solve it. Remember that the goal is to prove $\forall a > 0, b > 0$ there exists $p > 0, q > 0, r > 0$ such that $$\left(\frac{1}{p} + \frac{1}{q} + \frac{1}{r}\right)\left(\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{\frac{8}{ab} + 1}\right) < 4$$ Reducing the LHS (using Mathematica!), we get $$\frac{(8 p + 8 b p + a b p +...


6

You can show that $|\sin(x)|$ is subadditive, i.e. $$|\sin(x + y)| \le |\sin(x)| + |\sin(y)|.$$ To prove this, simply expand the left side: \begin{align*} |\sin(x + y)| &= |\sin(x)\cos(y) + \sin(y)\cos(x)| \\ &\le |\sin(x)| \cdot |\cos(y)| + |\sin(y)| \cdot | \cos(y)| \\ &\le |\sin(x)| + |\sin(y)|, \end{align*} as $|\cos(x)|$ and $|\cos(y)|$ are ...


7

Not sure that this is what you want, but a neat way to do it is noticing that if $0 < \theta < \pi$: $|1+e^{2i\theta}+...+e^{2i(n-1)\theta}|=\frac{|\sin (n\theta)|}{\sin (\theta)}$ and then use the triangle inequality on LHS


1

The inequality does always hold in a real inner product space. More generally, instead of $\|x-y\|,\|x-z\|\leq\frac12\|x\|$, we can require $\langle x,y\rangle,\langle x,z\rangle\geq\frac12\|x\|^2$. Lemma: For $a,b,c\in\mathbb{R}$ with $a\leq b$ and $b>0$ we have $$\frac{\sqrt{a^2+c^2}}{\sqrt{b^2+c^2}}\geq\frac{a}b.$$ Proof of lemma: If $a<0$, then ...


0

Many books refer to the operator $\log$ as in base $e$, although I do think that a proper notation is $\ln$, since this label has been specifically designed for the $e$ base. In order to find the reverse image in the considered interval you just evolve this inequality $$0<\ln{(x^2+2x+2)}<3$$ in this form $${x^2+2x+2}>1$$ $${x^2+2x+2}<e^3$$ ...


1

The Contradiction method works! Let $\frac{1}{\sqrt{a+1}}=p,$ $\frac{1}{\sqrt{b+1}}=q$ and $\frac{1}{\sqrt{c+1}}=r.$ Thus, $\{p,q,r\}\subset(0,1),$ $\frac{(1-p^2)(1-q^2)(1-r^2)}{p^2q^2r^2}=8$ and we need to prove that: $$p+q+r<2.$$ Indeed, let $p+q+r\geq2,$ $r=kr'$ such that $k>0$ and $p+q+r'=2$. Thus, $$p+q+kr'\geq2=p+q+r',$$ which gives $k\geq1.$ ...


0

You also know that $x^2\ge 0$, so you should write $0\le x^2\le 1$. Taking the square root is not enough, since $(-x)^2=x^2$. So if a positive $x$ satisfies the condition, so will $-x$. So if you take square toot, you get $$0\le x\le 1$$OR $$0\ge x\ge -1$$ Putting it all together, you get $$-1\le x \le 1$$


1

First of all, the direct image is wrong, it is not $(1,5]$ but $[1,5]$. Note $f(0)=1$ so $1$ also belongs to the image. As for the inverse image: you know that $x^2\ge 0$ anyways, so the condition is actually $0\le x^2\le 1$, i.e. $x\in[-1,1]$. Thus the inverse image is $[-1, 1]$. (I don't know why your book claims that it is $(-1,1)$, obviously $f(\pm 1)=2\...


0

After full expanding we need to prove that $$\sum_{cyc}(2a^3b-a^2b^2-a^2bc)\geq0$$ or $$\sum_{cyc}(a^3b+a^3c-a^2b^2-a^2bc)\geq\sum_{cyc}(a^3c-a^3b)$$ or $$\sum_{cyc}(a^3b+a^3c-a^2b^2-a^2bc)\geq(a-b)(b-c)(c-a)(a+b+c),$$ which is true because by Muirhead $$\sum_{cyc}(a^3b+a^3c-a^2b^2-a^2bc)\geq0$$ and by the given $$(a-b)(b-c)(c-a)(a+b+c)\leq0$$


0

Alternate answer to the $a_i = b_i$ case / too long for a comment. The $a_i = b_i$ case can also be solved by a greedy algorithm. Imagine the $a_i$'s as weights, and a balance where one arm is twice as long as the other arm. At any step: pick the heaviest remaining weight and put it on the side of the balance that is tilting up. If the balance is even (...


1

As an alternative, while not shorter, it can be brute-forced with just elementary algebra . . . Since the $\text{LHS}$ is homogeneous, we can assume $c=1$, and $a\ge b \ge 1$. Replacing $c$ by $1$, and simplifying, we get $$ 1-\text{LHS} = \frac {-a^2b+2a^3b-a^2-ab^2-a^2b^2-ab+2b^3-b^2+2a} {(a+b+1)(2a+b)(2b+1)(2+a)} $$ so it remains to show $$ -a^2b+2a^3b-...


1

Suppose $n\ge 10^{0.1}$. Then $x>1$ and $\log(\log n) \ge -1$. Therefore $$25^x>25x$$ $$10^{2x}>100(\frac{x}{4}2^{2x})\ge 100n$$ $$2x>\log n+2$$ $$2x>\log n-\log(\log n)+1.$$


3

The statement should have been $$(a^3+b^3)^2\leq (a^2+b^2)(a^4+b^4),$$ which is a result of the Cauchy- Schwarz inequality. Let $u=\langle a,\,b\rangle$ and $v=\langle a^2,\,b^2\rangle$. We have $$|u.v|^2\le ||u||^2||v||^2,$$ which implies the inequality.


1

Hint: $$(a^2+b^2)(a^4+b^4)-(a^3+b^3)^2={a}^{2}{b}^{2} \left( a-b \right) ^{2}\geq 0$$


1

The inequality is not true in the case when $a=b=0.5$. Indeed, $(0.5^2+0.5^2)(0.5^4+0.5^4)=\frac{1}{16}$ and $0.5^3+0.5^3=\frac{1}{4}$. However, the inequality$$(a^3+b^3)^2\leq (a^2+b^2)(a^4+b^4)$$always holds true for all real numbers $a$ and $b$. This comes using the CS inequality.


0

If there are no restrictions on $n$ other than $n\ge 1$, then the result is false. Let $x=2$ and $n$ tend to 1 from above.


1

Third time's the charm. I think it's false even when $z=x$. Let $(X,||\cdot||) = (C([0,1]),||\cdot||_\infty)$. We have $||x||_\infty = 1, ||y||_\infty = \frac{3}{4}, ||x-y||_\infty = \frac{1}{2}$, and $||(1+\frac{1/2}{3/4})y+x||_\infty = ||\frac{5}{3}y+x||_\infty = \frac{11}{6} < 2 = 2||x||_\infty$.


2

There is the generalized mean value theorem which says that for all intervals $(a,b)$ in the intersection of the domains of a pair of differentiable functions $f$ and $g$ there exists $c \in (a,b)$ such that $$ f'(c)(g(b) - g(a)) = g'(c)(f(b) - f(a)) $$ So $$ |f'(c)||g(b) - g(a)| = |g'(c)||f(b) - f(a)| \geq |f'(c)||f(b) - f(a)| $$ So either $f'(c) = 0$ or $$ ...


2

Counterexample: $n:=2 ~ , ~~\displaystyle x:=-\frac{1}{2} :$ $\displaystyle \frac{1}{2}-\frac{x}{4}+\frac{x^2}{8} = 0.65625$ $\displaystyle \left(\frac{1}{7-7x^3}\right)^{1/3} = 0.5026316274194358675807...$ $n\to\infty$ : $7-7x^3 \leq (x+2)^3$ Counterexample with $~\displaystyle x:=-\frac{1}{2}~$: $7-7x^3 = 7.875$ $(x+2)^3 = 3.375$


0

Before we start, i would like to insert some comments showing why this problem is a "difficult problem", and why we need for it rather ad-hoc methods of attack. To give full details of the computations, i must also use computer aid, but i hope that dryly accepting the results is compensated by the obtained insight. As it often happens with inequalities with ...


1

Since it seems you are confused, let me include a self-contained proof that does not involve Jensen's inequality, to make it crystal clear why this is true. Consider the quantity which is obviously non-negative: $$ \mathbb E\bigl(Y^2-\mathbb E(Y^2)\bigr)^2\geq 0. $$ Now expand the square and collect like terms to learn that $$ \mathbb E\bigl(Y^2-\mathbb E(Y^...


3

If $f'$ is integrable, you can say $$ \lvert f(x)-f(y) \rvert = \left\lvert \int_y^x f' \right\rvert \leq \int_y^x \lvert f' \rvert \leq \int_y^x \lvert g' \rvert = V_y^x(g) , $$ the total variation of $g$. You can see this in play in a pair with $g'$ taking on positive and negative values, $f' = \lvert g' \rvert$. Then $f$ can get quite large while $g$ ...


0

Note that $xyz(x-y)(x-z)(y-z)$ is cyclic. WLOG, assume that $z = \min(x, y, z)$. We only need to prove the case when $(x-y)(x-z)(y-z) > 0$. In other words, we only need to prove the case when $x > y > z$. Let $$A = 2 + 2\cos \frac{\pi}{9}, \quad B = 2 - 2\cos \frac{4\pi}{9}, \quad C = 2 - 2\cos \frac{2\pi}{9}.$$ Clearly $A > B > C > 0$. ...


0

Hint: Note that we can rewrite $\;S_n=2-\dfrac 5{n+3}$, which shows $(S_n)$ is an increasing sequence which converges to $2$. Do, as $2-1.99 <0.1$, to ensure that $|S_n-1.99|<0.1$, it suffices to choose $n$ such that $$ S_n>1.99,\enspace\text{i.e. }\;\frac 5{n+3}<0.01.$$


2

Yes it is true that $$|S_n-1.99|\lt 0.1$$ for $n\gt42$. But we cannot say that $$\forall\epsilon\gt0\quad\exists N:\forall n\gt N\quad|S_n-1.99|\lt \epsilon$$ For example take $\epsilon=0.005$ and we have that $\forall n\gt996$ the inequality is false. Hence one cannot say that $$\lim_{n\to\infty}S_n=1.99$$


0

Late answer since it was tagged as duplicate there and this way of solving is neither here nor there. \begin{eqnarray*} abc(a+b+c) & \stackrel{GM-AM}{\leq} & \left(\frac{a+b+c}{3}\right)^3(a+b+c)\\ & = & 3\left(\frac{a+b+c}{3}\right)^4 \\ & \stackrel{x^4 \; is \; convex}{\leq} & 3\cdot \frac{1}{3}(a^4+b^4+c^4) = a^4+b^4+c^4 \end{...


0

\begin{eqnarray*} &(x^2-y^2)^2+(y^2-z^2)^2+(z^2-x^2)^2 \\&+2(z^2-xy)^2 +2(x^2-yz)^2+2(y^2-zx)^2\geq 0 \end{eqnarray*} now divide by $4$ and rearrange.


2

By AM_GM we get \begin{align*} \frac{2a^4+b^4+c^4}{4} & \geq a^2bc\\ \frac{2b^4+a^4+c^4}{4} & \geq b^2ac\\ \frac{2c^4+b^4+a^4}{4} & \geq c^2ab \end{align*} Now add these to get your inequality.


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