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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

$$\frac { a_{ n+1 } }{ a_{ n } } =\frac { \left( 1+\frac { 1 }{ n+1 } \right) }{ { \left( 1+\frac { 1 }{ n } \right) }^{ n } } ^{ n+1 }=\left( 1-\frac { 1 }{ (n+1)^{ 2 } } \right) ^{ n+1 }\frac { n+1 }{ n } >\left( 1-\frac { 1 }{ n+1 } \right) \frac { n+1 }{ n } =1$$


3

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 that $$1\cdot (2n-1)<n^2$$ $$3 \cdot (2n-3)<n^2$$ $$\ldots$$ $$\ldots$$ 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 ...


2

$$1=(a+b+c+d+e)^2 \geqslant \sum_{cyc} a^2 + 2S$$ 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 $$e^z - 1 = z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots$$ and 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 $...



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