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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. $$\frac {(4\alpha x^2)+2(\frac 1 {2x})} {3} \ge \sqrt[3]{ (4\alpha x^2)\Big(\frac 1 {2x}\Big)^2}$$ See if you can proceed now.


2

$$4\alpha x^2+\frac{1}{2x}+\frac{1}{2x}\geq 3\sqrt[3]{4\alpha x^2\cdot \frac{1}{2x}\cdot \frac{1}{2x}}$$


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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