# Tag Info

11

Note that $$5a^2+b^2+c^2-4ab-2ac=(2a-b)^2+(a-c)^2\ge 0.$$

7

Hint: $$\frac{1}{x-6} \leq 3 \iff \frac{1}{x-6} - 3 \leq 0 \iff \frac{19-3x}{x-6} \leq 0 \iff \frac{3x-19}{x-6} \geq 0$$ Can you take it from here?

5

Your first step is wrong. If $A\le 3$, that doesn't mean $\dfrac 1 A\le \dfrac 1 3$. It does mean $\dfrac 1 A \ge \dfrac 1 3$ if $A$ is positive, but we can't assume it's positive in this case. If you write this as $\dfrac 1 {x-6} - 3\le 0$, then use a common denominator to get just one fraction, then do the routine simplifications, you'll be well on your ...

4

One may see that for the initial inequality to hold true it is sufficient to prove that $$1-\frac{(2-x) (3-x) \left(36-84 x+89 x^2-50 x^3+13 x^4\right)}{16 x(1-x)^2 }<0,\quad x \in (0,1), \tag1$$ then setting \begin{align} &f(x)=16 x(1-x)^2-(2-x) (3-x) \left(36-84 x+89 x^2-50 x^3+13 x^4\right) \end{align} one gets\begin{align} &f'(x)=... 3 The GRE question asks: If y<0, which of \frac{1}{y + \frac{1}{y}} and y is greater. This is not a numerical evaluation question (unless the answer is "cannot be determined"), but rather an algebraic reasoning question. Since y is negative, when y is added to \frac{1}{y} we get a number that is less than \frac{1}{y}. That is, y + \frac{1}{... 3 Let n=3. Hence, by AM-GM we get x_1x_2x_3+x_2x_3x_2+x_3x_1x_2\leq3\left(\frac{x_1+x_2+x_3}{3}\right)^3=\frac{1}{9}. The equality occurs for x_1=x_2=x_3=\frac{1}{3}. Id est, the answer is \frac{1}{9}. Let n=4. Hence, by AM-GM x_1x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2= =x_1x_2(x_3+x_4)+x_3x_4(x_1+x_2)\leq\left(\frac{x_1+x_2}{2}\right)^2(x_3+... 3\frac { a_{ n+1 } }{ a_{ n } } =\frac { \left( 1+\frac { 1 }{ n+1 } \right) }{ { \left( 1+\frac { 1 }{ n } \right) }^{ n } } ^{ n+1 }=\left( 1-\frac { 1 }{ (n+1)^{ 2 } } \right) ^{ n+1 }\frac { n+1 }{ n } >\left( 1-\frac { 1 }{ n+1 } \right) \frac { n+1 }{ n } =13 What if you fix \epsilon>0 and consider f:[0,3]\rightarrow\mathbb{R} defined by f(x) = (x-1)^2 + \epsilon? So m=0, M=3. Let x_1=0, x_2=2. So: \begin{align} \frac{f(x_1)+f(x_2)}{2} &= \frac{f(0)+f(2)}{2} = 1 + \epsilon\\ f(\frac{x_1+x_2}{2}) &= f(1) = \epsilon \\ \frac{f(M)+f(m)}{2} &= \frac{f(3)+f(0)}{2} = (5/2)+\epsilon\\ f(\... 3 Hint: a<b\iff\dfrac1a>\dfrac1b is true only if a and b have the same sign. Hence you have to consider two cases. 3 Non-Inductive Proofs: (1) Observe that1\cdot (2n-1)<n^23 \cdot (2n-3)<n^2\ldots\ldots$$In general,$$r\cdot(2n-r) <n^2$$Multiplying all these inequalities we get the desired result, when n is even. When n is odd, we need to multiply both sides by n to get the desired result. (2) Just as Zain Patel commented: ... 3 If x < 6, then x - 6 is negative, so multiplying both sides by x-6 reverses the inequality. Thus 1 \ge 3(x - 6), or 19 \ge 3x, which is true for all x < 6. 3 Using the mL (minumum times Length) estimate$$\int_a^b |f(x)|dx\geq (b-a)\min_{x\in[a,b]}|f(x)|$$One has$$\int_1^n\frac{1}{x}dx=\sum_{k=1}^{n-1}\int_{k}^{k+1}\frac{1}{x}dx\geq\sum_{k=1}^{n-1}(k+1-k)\frac{1}{k+1}=\sum_{k=1}^{n-1}\frac{1}{k+1}=\sum_{k=1}^n\frac{1}{k}-1,where I have used that \frac{1}{x} is positive on the positive axis (so the absolute ... 3 \begin{align*}9\cos^2x−10\cos x\sin y−8\cos y\sin x+17 &= 9\cos^2x−2\cos x\sin y−8(\cos x\sin y +\cos y\sin x)+17 \\ & = 9\cos^2x−2\cos x\sin y−8\sin(x+y)+17 \\ & \geq -2-8+17 \\ & \geq 7 \\ & \geq 1 \end{align*} Admittedly my confidence in this answer is not very high. EDIT: I plugged the equation into wolfram alpha and told it to ... 2 They are tight, in the sense that we have \text{trace}(AB) = \lambda_{max}(A)\; \text{trace}(B) if A = I. Similarly in the second one if B=I. 2 I remember encountering the same confusion as you when presented with such inequalities in optimization problems. x = (x_1, \dotsc, x_n)^T is a column vector of variables which are usually under our control in the problem, but subject to certain constraints. For instance, you might encounter the constraint x \geq 0. In the vector terms you may be used ... 21=(a+b+c+d+e)^2 \geqslant \sum_{cyc} a^2 + 2S$$Further from CS inequality$$(1+1+1+1+1) \cdot\sum_{cyc} a^2 \geqslant \left(\sum_{cyc} a \right)^2=1$$Thus 1 \geqslant \frac15+2S \implies S \leqslant \frac25  2 Notice that: 4a^2+b^2+a^2+c^2 \geq 4ab+2ac By AM-GM inequality. EDIT: The reason why I come up with such idea is because b and c in RHS is independent. So ab comes from one AM-GM and ac comes from the other. Notice that 4ab = 2\sqrt{4a^2b^2}  since the 2 is the constant deriving from AM-GM, one shall find the suitable combination of ... 2 The general rule for inverses in inequalities is this: If a and b have the same sign, \;a<b\iff \dfrac1b<\dfrac1a. 2 The left hand side is$$\frac1{(n+1)^3}\cdot((n+1)^2-n)(n+2)=\frac{n^3+3n^2+3n+2}{(n+1)^3} $$2 Let$$f(n) = \frac{1}{2^2} + \cdots + \frac{1}{n^2}.$$Now we want to show that$$f(n)<\frac{n-1}{n}\tag{1}$$for all integers greater than 1. A proof by induction consists of two equally important steps. In the base case we show that (1) indeed holds for n=2. In the inductive step we assume that (1) is true for some number n and use that to ... 2 You made a second conjecture in the comment to my first answer, so I am answering that here. The conjecture is: Conjecture:$$ \frac{f(x_1)+...+f(x_n)}{n} - f(\frac{x_1+...+x_n}{n}) \leq \frac{f(M)+f(m)}{2}- f(\frac{M+m}{2}) $$whenever f is continuous and convex over the interval [m,M] and m \leq x_i\leq M for all i. Counter-example: Consider ... 2 Can I do that for least value of a, the value of 3\sqrt[3]{a}  must greater than minimum value of 4ax^2 + \frac{1}{x} I find this strange because "minimum value of 4ax^2 + \frac{1}{x}" is 3\sqrt[3]{a}. So I guess that you meant that "the value of 3\sqrt[3]{a}  must greater than 1". Then, it is correct. I'll write the details in the ... 2 Note that if a,b,c are all equal to \frac{1}{2}, then the given products are all \frac{1}{4}. So it is not true that at least one of a(1-b), b(1-c), and c(1-a) is \lt \frac{1}{4}. But we can prove that at least one is \le \frac{1}{4}. If one or more of 1-a, 1-b, 1-c is \le 0, the result is obvious. So suppose they are all positive.... 2 To complete the given answer by user11235813:$$\frac{3x-19}{x-6} \geq 0\to 3x\geq 19\ \&\ x\geq 6 \text{ or } 3x\leq 19 \ \&\ x\leq 6$$So, the answer is$$x\geq \frac{19}3\ \cup\ x\leq 6$$that is$$\mathbb{R}-(6,\frac{19}3)$$2 With one variable, simplest is cases. If x=6 it is not defined. If x<6, as you noted LHS is negative, and the inequality trivially holds. If x>6, we may multiply throughout by the positive quantity x-6, and then it should be easy. In the end take the Union of the allowable regions you get. 2 Start with the series expansion$$e^z - 1 = z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdotsand use the triangle and reverse triangle inequalities to get the result. By the triangle inequality and the condition \lvert z\rvert \le 1, we have\begin{align}\lvert e^z - 1\rvert &\le \lvert z \rvert + \frac{\lvert z\rvert^2}{2!} + \frac{\lvert z\rvert^... 2 Since y \neq 0 it is justified to multiply the numerator and denominator of a fraction by y, without changing the value of the fraction.\frac{1}{y+\frac{1}{y}} = \frac{y}{y^2+1}$$Now compare this to y by subtracting y and comparing to zero:$$\frac{y}{y^2+1}-y = \frac{y-y^3-y}{y^2+1} = -\frac{y^3}{y^2+1} $$We will not change the sign of this ... 2 Here is a slicker proof for Question 1: The Frobenius rank inequality says that if m,k,p,n are four nonnegative integers, and if U\in\mathbb{F}^{m\times k}, V\in\mathbb{F}^{k\times p} and W\in\mathbb{F}^{p\times n} are three matrices, then \operatorname*{rank}\left( UV\right) +\operatorname*{rank}\left( VW\right) \leq\operatorname*{rank}V+\... 1 At the beginning we transform the original inequality:$$\dfrac{8x^4}{8x^3+5y^3}+\dfrac{8y^4}{8y^3+5z^3}+\dfrac{8z^4}{8z^3+5x^3}\geq \dfrac8{13}(x+y+z),x-\dfrac{5xy^3}{8x^3+5y^3}+y-\dfrac{5yz^3}{8y^3+5z^3}+z-\dfrac{5zx^3}{8z^3+5x^3}\geq \dfrac8{13}(x+y+z),\dfrac{xy^3}{8x^3+5y^3}+\dfrac{yz^3}{8y^3+5z^3}+\dfrac{zx^3}{8z^3+5x^3}\leq \dfrac1{13}(x+y+z)....

1

Formally, $x,y,z>0$, but inequality has a continuous extension also at the edges of the field. Let $z=0$ for definiteness, then the problem transforms to $$\dfrac{xy}{5y^3+4}\leq \dfrac13,\quad x+y=3,$$ $$\dfrac{(3-y)y}{5y^3+4}\leq\dfrac13,$$ $$h(y) = 5y^3+3y^2-9y+4\geq 0.$$ Taking in account that $$h'(y)=15y^2+6y-9 = 3(5y-3)(y+1),$$ the minimal value of \$...

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