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## New answers tagged inequality

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 Letx_0=0$. Thus, by C-S $$\sum_{k=1}^n\frac{x_k}{1+\sum\limits_{i=1}^kx_i^2}\leq\sqrt{n\sum_{k=1}^n\frac{x_k^2}{\left(1+\sum\limits_{i=1}^kx_i^2\right)^2}}\leq$$ $$\leq\sqrt{n\sum_{k=1}^n\frac{x_k^2}{\left(1+\sum\limits_{i=0}^{k-1}x_i^2\right)\left(1+\sum\limits_{i=1}^kx_i^2\right)}}=\sqrt{n\left(\frac{\sum\limits_{k=1}^nx_k^2}{1+\sum\limits_{k=1}^nx_k^2}\... 0 If h is not equal to zero almost everywhere, then \alpha(h)=1. First let us restrict ourselves to f,g \in L^2(\mathbb{R}). Otherwise we run into troubles with multiplying zero and infinity. I am assuming that we are talking about the Lebesgue measure with the usual Lebesgue sigma-algebra. Case 1: h equal to zero almost everywhere. Let's take care of ... 2 Hint: apply A.M >G.M between for (1-a)(1-b)(1-c) \sqrt{(1-a)(1-b)(1-c)} \le \cfrac{(1-a)+(1-b)+(1-c)}{3}$$ 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 > 0such 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 Supposen\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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