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## Hot answers tagged inequality

5

Clearly $x\geq -2$. If $x<0$ then each $x\in [-2,0)$ is a solution (since negative number is always smaller than square root). Now if $x\geq 0$ then you can square it, so you get $$x^2-x-2 = (x-2)(x+1)\leq 0$$ So in this case every $x\in[0,2]$ is a solution. So finally, every $x\in [-2,2]$ is a solution.

4

Once you know $x \geqslant -2$, consider first $x \in [-2, 0)$. The LHS is defined and non-negative, while the RHS is __________. Next, consider the case $x \geqslant 0$, where you can freely square as you have done. Here you should get $x \in [0, 2]$ as the solution. Now the solution set is the union of these cases.

4

Be careful : When you square, the inequality preserves its sign direction if both sides are positive. Note that $\sqrt{x+2}$ is defined for $x \geq - 2$, so first you need to consider $x \geq 0$ and work as such : $$\sqrt{x+2} \geq x \Rightarrow x+2 \geq x^2 \Leftrightarrow x^2-x-2 \leq 0 \Leftrightarrow (x-2)(x+1) \leq 0$$ This indeed yields $x \in [-1,2]... 4 Introduce: $$1+a=x^2,\ \ 1+b=y^2$$ Obviously: $$2\le x<y\le3$$ Notice that: $$4\le y + x \le 6\tag{1}$$ Inequality now becomes: $$\frac{y^2-x^2}{6}\le y-x\le\frac{y^2-x^2}{4}$$ $$\frac{y+x}{6}\le 1\le\frac{y+x}{4}$$ ...which is true because of (1). 4 You can’t tell; the most you can say is that x is between a and d. Also i don’t see how point e is significant. One way to think about what it means geometrically is to take the center of gravity of the points; imagine putting 4 equal sized weights at a b c d on a number line. 4 By the weighted AM-GM inequality,$a^ab^bc^c \leq a \times a + b \times b + c \times c$(by using weights$a$,$b$,$c$). Similar expressions hold for the other terms. So the inequality becomes$a^ab^bc^c + a^bb^cc^a + a^cb^ac^b \leq a^2 + b^2 + c^2 + ab + bc + ca + ac + ba + cb = (a+b+c)^2 = 1.$3 Case$i = K$is trivial, so we can assume$S_i < 1$.$P_K = P_i \cdot \prod_{j=i+1}^K (1 - a_j)$and$\sum_{j=1+1}^K a_j = 1 - S_i$, we have$P_K \geqslant P_i \cdot (1 - a_{i + 1} - a_{i + 2}) \cdot \prod_{j=i+3}^K \ldots \geqslant P_i \cdot(1 - a_{i + 1} - \ldots - a_K) = P_i \cdot S_i$and so$1 - P_K \leqslant 1 - P_i \cdot S_i$, so it's enough to ... 3 An alternative approach using calculus: when$\alpha<0,4\alpha x^2+\frac1x\to-\infty<1$as$x\to\infty.\alpha=0$may be rejected similarly. Thus,$\alpha>0$.$4\alpha x^2+\frac1x$is differentiable for$x>0$and attains global minimum of$3\alpha^{1/3}$at$x=\frac1{2\alpha^{1/3}}$. Thus,$$3\alpha^{1/3}\ge1\\\implies\alpha\ge\frac1{27}$$ 3 The limit is$\|f\|_{\infty}=\sup \{|f(x): 0\leq x \leq 1\}$. It is clear that$(\int|f|^{n})^{1/n} \leq \|f\|_{\infty}$. Now there exists$a$such that$|f(a)| =\|f\|_{\infty}$. By continuity given$\epsilon >0$there exists$\delta >0$such that$|f(x)| >\|f\|_{\infty}-\epsilon$for$|x-a| <\delta$. Hence$(\int|f|^{n})^{1/n} \geq (\int_{(a-\...

3

Try to use weighted AM-GM inequality: for any $x,y,z,p,q,r>0$ with $p+q+r=1$ one has $$x^py^qz^r \le px+qy+rz.$$

3

Let $\alpha = -\frac{3}{2} + \frac{\sqrt{3}}{2}i$. We have $z_{n+1}-z_n = \alpha^n (\alpha-1)$, and so $|z_{n+1}-z_n| = |\alpha^n (\alpha-1)| = |\alpha|^n \cdot|\alpha - 1|$, since multiplying complex numbers multiplies their modulus. You now solve $|\alpha|^n \cdot|\alpha - 1| > \sqrt{7000}$ which after computing $|\alpha|$ and $|\alpha-1|$, and then ...

2

One can in fact show that $$\frac{1-P_1}{S_1} > \frac{1-P_2}{S_2} > \ldots > \frac{1-P_K}{S_K} \, .$$ if all $a_i \in (0, 1)$. If, in addition, $\sum_{i=1}^{K} a_i = 1$ then the last term is equal to $1-P_K$, and the desired conclusion follows. Proof: For $1 \le i \le K-1$ \begin{align} \frac{1-P_i}{S_i} - \frac{1-P_{i+1}}{S_{i+1}} &= \... 2 The relation is not symmetric. For example we have for a=0 and b=3: a^2-b^2 =-9 \le 7, but b^2-a^2=9 > 7. 2 Because \sqrt2 is irrational, 2n^2-m^2\ge1 or 2n^2-m^2\le-1. In the first case,(\sqrt2n-m)(\sqrt2n+m)\ge 1,$$so$$\sqrt2-\dfrac mn\ge\dfrac1{n(\sqrt 2 n+m)} \ge\dfrac 1 {n(\sqrt2n+\sqrt2n)}\ge\dfrac1{2\sqrt2 n^2}\ge\dfrac1{3n^2}.$$In the second case, m^2-2n^2\ge1, so (m-\sqrt2n)(m+\sqrt2n)\ge1, so \dfrac mn-\sqrt2\ge\dfrac1{n(m+\sqrt2n)}... 2 Depending on the precise definition of the order, addition, and the properties you know, you can argue as follows: You have b=a+d for some d, with d\gt 0. Since d\gt 0, it is a successor, so d=c\mathrm{++} for some c\geq 0. Thus,$$b = a+d = a+(c\mathrm{++}) = (a+c)\mathrm{++} \geq a\mathrm{++}$$since a+c\geq a. 2 Let a=|x|^{a_1},b=|y|^{a_2}, p=\frac{b_1}{a_1}, q=\frac{b_2}{a_2}, p,q>1, \frac{1}{p}+\frac{1}{q}=1, |x|^{b_1}+|y|^{b_2}=a^p+b^q Young inequality is ab \le \frac{1}{p}a^p+\frac{1}{q}b^q, but p,q>1 imply \frac{1}{p}a^p+\frac{1}{q}b^q<a^p+b^q, so putting things together we get \frac{|x|^{a_1}|y|^{a_2}}{|x|^{b_1}+|y|^{b_2}}=\frac{ab}{a^p+b^... 2 Clearly you can see, x does not cancel on RHS You need to use the weighted AM-GM inequality.$$\frac {(4\alpha x^2)+2(\frac 1 {2x})} {3} \ge \sqrt{ (4\alpha x^2)\Big(\frac 1 {2x}\Big)^2}$$See if you can proceed now. 2$$4\alpha x^2+\frac{1}{2x}+\frac{1}{2x}\geq 3\sqrt{4\alpha x^2\cdot \frac{1}{2x}\cdot \frac{1}{2x}}$$2 The least value is 1/27. Here is why. Suppose that \alpha\in\mathbb{R} satisfies$$ (\forall x>0)\quad 4\alpha x^2+\frac{1}{x}\geq 1. $$We therefore must have (\forall x>0)\;\alpha\geq(x-1)/(4x^2). A straightforward calculation shows that the maximum value of the function x\mapsto(x-1)/(4x^2) on \left]0,+\infty\right[ is 1/27. Thus \... 2 Note the x^2 in the denominator ensures the function is not defined at x=0, otherwise has no effect on the inequality, so we may ignore it. Similarly, \sin x -2 is always negative, and so is -x^2+x-2, so both may be together ignored. As e^x-1 has the same sign as x, essentially we can substitute that, and equivalently solve for$$\frac{x(x-1)(x-...

2

It is true if $A$ and $B$ are real. For symmetric real matrices, the spectral radius agrees with the operator norm $$\|A\|=\max\{\|Ax\|_2:\ \|x\|_2=1\}.$$ And the sum of real symmetric matrices is real symmetric. Thus $$\rho(A+B)=\|A+B\|\leq\|A\|+\|B\|=\rho(A)+\rho(B).$$ Non-negative is not necessary for all the above. The same result holds for ...

2

Hint: $5^n > 2^{2n+1}$ for $n \ge \ldots$.

1

As pointed out in the comments the inequality: $$\large \frac{1}{AM \cdot BN} + \frac{1}{BN \cdot CP} + \frac{1}{CP \cdot AM} \le \frac{4}{3(R - OI)}$$ Is not homogeneous and therefore cannot be correct. Take any triangle and any point and even if the given inequality is satisfied fot this configuration then after scaling it by $a$ for sufficiently small $a$ ...

1

A reasonable option is $$\{(x,y)\in\Bbb R^2\,:\, y>x^2-4\}$$

1

Since $\lambda=\Re\lambda+i\Im\lambda$, we solve $$|1-\tau\lambda|=|(1-\tau\Re\lambda)-i\tau\Im\lambda|=\sqrt{(1-\tau\Re\lambda)^2+(\tau\Im\lambda)^2}<1.$$ Squaring both sides and tidying yields $$1-2\tau\Re\lambda+\tau^2|\lambda|^2<1\implies\tau(|\lambda|^2\tau-2\Re\lambda)<0.$$ Thus $$0<\tau<\frac{2\Re\lambda}{|\lambda|^2},$$ and note that $-... 1 I don't think it is a mistake. The inequality system$\displaystyle \begin{cases}|ax+by|\le c \\|dx+ey|\le f \end{cases}$(where$c,f>0$) defines a region in the shape of a parallelogram, with sides$ax+by=\pm c$and$dx+ey=\pm f$. The regions defined by$\displaystyle \begin{cases}|x|\le k_1 \\|y|\le k_2 \end{cases}$is a rectangle and The regions ... 1 In style to the Liouville's theorem mentioned in the comments,$\sqrt{2}$is a root of$P_2(x)=x^2-2$. Then, for any$\frac{m}{n}$we have an$\varepsilon$in between$\sqrt{2}$and$\frac{m}{n}$such that (this is MVT) $$\left|P_2\left(\frac{m}{n}\right)\right|= \left|P_2(\sqrt{2})-P_2\left(\frac{m}{n}\right)\right|= |P_2'(\varepsilon)|\cdot \left|\sqrt{2}-\... 1 Hint: The inequation \;\sqrt A\ge B, on its domain (defined by the condition A\ge 0) is equivalent to$$A\ge B^2\quad\textbf{ or }\quad B\le 0.$$1 Let a=\sqrt{x+2}\ge0 for real x We need$$a\ge a^2-2\iff0\ge a^2-a-2=(a-2)(a+1)\iff -1\le a\le2\ \ \ \ (1)$$But we need to honor a\ge0\ \ \ \ (2) Find the intersection of (1),(2) 1 It suffices to show that$$4(\,x+ y+ z\,)^{\,3}- 27(\,xy^{\,2}+ yz^{\,2}+ zx^{\,2}+ xyz\,)\geqq 0$$Consider now x\equiv \text{mid}\{\,x,\,y,\,z\,\} Let F= 4(\,x+ y+ z\,)^{\,3}- 27(\,xy^{\,2}+ yz^{\,2}+ zx^{\,2}+ xyz\,) Thus, we have$$F= (\,x+ 4\,z\,)(\,y+ z- 2\,x\,)^{\,2}+ 4\,y(\,y- z\,)^{\,2}- 11\,y(\,x- y\,)(\,x- z\,)\geqq 0$\$

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