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14

Let $a=\sqrt{x}+\sqrt{x+7}$, then $a+a^2<42$, and $-7<a<6$


9

Let $n \in \mathbb{N}.$ You may write $$ 4^n=2^{2n} =(1+1)^{2n}= \sum_{k=0}^{2n}\binom{2n}{k}\geq\binom{2n}{n}. $$


8

$\displaystyle 4^n=(1+1)^{2n} =\sum_{k=0}^{2n}\binom{2n}{k}\geq\binom{2n}{n}$


8

A shorter proof would note that we always have $y \ge \lfloor y \rfloor$, hence ${x \over n} \ge \lfloor {x \over n} \rfloor$. Multiplying by $n$ gives $x \ge n\lfloor {x \over n} \rfloor$. Taking the floor of both sides gives the desired result (noting that the floor of an integer is the integer).


7

A bit more can be proven with a bit more work. For $k\ge0$, we have the inequality $$ \begin{align} \left(\frac{k+\frac12}{k+1}\right)^2 &=\frac{k^2+k+\frac14}{k^2+2k+1}\\ &\le\frac{k+1}{k+2}\tag{1} \end{align} $$ because cross-multiplication gives $k^3+3k^2+\frac94k+\frac12\le k^3+3k^2+3k+1$. Using $(1)$ yields $$ \begin{align} ...


5

$$a+b> \sqrt{a^2+b^2-ab} \iff (a+b)^2>\left (\sqrt{a^2+b^2-ab}\right )^2$$ $$ \iff\quad a^2+2ab+b^2>a^2+b^2-ab$$ $$\iff\quad 3ab>0 \quad \iff\quad ab>0$$ which is correct because of the hypothesis $a,b >0$.


4

In fact, we only need that $x \geq 1$: \begin{align*} x \geq 2 &\implies x \geq 1 \\ &\implies x^6 \geq 1^6 \\ &\implies \frac{1}{x^6} \leq \frac{1}{1^6} \\ &\implies \frac{4}{x^6} \leq \frac{4}{1^6} \\ &\implies 1 + \frac{4}{x^6} \leq 1 + \frac{4}{1^6} \\ &\implies \sqrt{1 + \frac{4}{x^6}} \leq \sqrt{1 + \frac{4}{1^6}} = \sqrt{5} \\ ...


4

We have $\frac{ab^2}{a^2 +b^2}\leq \frac{b}{2} ,\frac{bc^2}{c^2 +b^2}\leq \frac{c}{2} ,\frac{ca^2}{c^2 +a^2}\leq \frac{a}{2}$ hence $$\frac{a^3}{a^2 +b^2} +\frac{b^3}{c^2 +b^2} +\frac{c^3}{a^2 +c^2} =a+b+c - \left(\frac{ab^2}{a^2 +b^2}+\frac{bc^2}{c^2 +b^2}+\frac{ca^2}{c^2 +a^2}\right) \geq \frac{a+b+c}{2}.$$


4

You can get a much more convenient condition by using $(a-r)(b-r)=ab-ar-br+r^2$ for a value of $r$ other than $1$.


4

You can prove that $2 | \langle a, b \rangle | \leq \| a \|^2 + \| b \|^2$ as I believe you have proven to attain $\| a + b \|^2 \leq 2\| a \|^2 + 2\| b \|^2$. Then you just expand: $\| a + b + c \|^2 = \langle a, a \rangle + \langle b, b \rangle + \langle c, c \rangle + 2\langle a, b \rangle + 2\langle a, c \rangle + 2\langle b, c \rangle \leq \| a \|^2 + ...


3

You can expand the LHS and use $\left<x,y\right>+\left<y,x\right>\leq ||x||^2+||y||^2$ 3 times. This latter inequality follows by expanding the non-negative entity $\left<x-y,x-y\right>$.


3

Do you know Schur's Inequality,i.e. If $x,y,z\geqslant 0$, then $$x(x-y)(x-z)+y(y-x)(y-z)+z(z-x)(z-y)\geqslant 0.$$ Proof: Without loss of generality, we can assume that $x\geqslant y\geqslant z$. Then $$x(x-y)(x-z)+y(y-x)(y-z)\geqslant x(x-y)(x-z)+x(y-x)(y-z)=x(x-y)^2\geqslant 0$$ and $$z(z-x)(z-y)\geqslant ...


3

OK. Let me try to complete the answer. We want to prove that for $a,b,c\in \mathbb{R},abc>0$, $$ab+bc+ac<\dfrac{\sqrt{abc}}{2}+\dfrac{1}{4}. \tag{1}$$ In a previous post, it is proved for $a,b,c\ge 0$ Because $abc>0$, the only other possibility is that two of the numbers are negative and one is positive. We can assume that ...


3

Let $g(x) = \displaystyle \int_0^x \frac{h(t)}{f(t)}dt$, where we will select $h>0$ later. Then $g' = \displaystyle \frac {h}{f}$ and $g'' = \displaystyle \frac {fh'-hf'}{f^2}$. So you get $$f'g'+2fg'' = f'\frac{h}{f}+2\frac{fh'-hf'}{f} = \frac{2fh'-hf'}{f}$$ For this to be positive, we need $2f h' > h f' \iff \dfrac{h'}{h} > \dfrac{f'}{2f}$. So ...


3

A start: Square everything. The inequality on the right will be obvious. The one on the left is a little more unpleasant.


3

Your first implication $2\leq x\implies \frac{1}{x}\geq\frac{1}{2}$ is actually false: you have to flip the inequality direction, roughly because taking the reciprocal of a large number will give you a small number. Here is how you can do it: \begin{align*} 2 &\leq x \\ \implies \frac{1}{x} &\leq \frac{1}{2} \\ \implies \frac{1}{x^6} &\leq ...


3

Since $p>1$ and $x,y>0$ then $$\left(\frac{x}{x+y}\right)^p<\frac{x}{x+y}\textrm{and }\left(\frac{y}{x+y}\right)^p<\frac{y}{x+y}.$$ It follows that $$\left(\frac{x}{x+y}\right)^p+\left(\frac{y}{x+y}\right)^p<\frac{x}{x+y}+\frac{y}{x+y}=1$$ that leads to what you wanted.


3

Let $x\neq0$ $1>1-\frac{1}x\Leftrightarrow 0>-\frac{1}x$ If $x\in\mathbb{N},x\neq0$ then $x>0$ hence $\frac{1}x>0$, therefore the inequality always hold


3

Let $f(x)=x+\cos x-1$. We have $f(0)=0$. Show that $f'(x)\ge 0$ for $x\gt 0$. It will follow that $f$ is non-decreasing on the interval $[0,\infty)$. Note that $f'(x)\gt 0$ in, say, the interval $[0,\pi/6)$. So $f$ is strictly increasing in this interval, and is therefore positive for all $x\gt 0$. (Actually, $f$ is strictly increasing on the whole interval ...


3

By Taylor's theorem with just one term (you can't use 2 terms as you have no information on if $f''$ exists), for any $x\in [0,2]$ we have $$\begin{align}f(x) &= f(0) + f'(c)x \\ f(x) &= f(2) + f'(d)(x-2)\end{align}$$ for some $c,d \in [0,2]$. The first one shows $$1-x \leq f(x) \leq 1 + x \qquad (*)$$ and the second shows $$1 +(x-2) \leq f(x) \leq ...


3

Note that $2ab > -ab \Rightarrow a^2+b^2+2ab > a^2 + b^2 -ab \Rightarrow (a+b)^2 > a^2 + b^2 -ab \Rightarrow a+b > \sqrt{a^2 + b^2 -ab}$ I was able to do this by squaring the original inequality I was supposed to prove, and then I worked backwards.


3

$3ab>0$ Hence, $2ab>-ab$ Hence, $a^2+2ab+b^2>a^2+b^2-ab$ Hence $a+b>\sqrt{a^2+b^2-ab}$ because $a^2+b^2-ab>a^2+b^2-2ab=(a-b)^2>0$


2

Using AM-GM, we have: $$4x^2+9y^2 \ge 12xy$$ $$3y^2+12z^2 \ge 12yz$$ $$18z^2+2x^2\ge 12zx$$ Combining and using the constraint, $$6x^2+12y^2+30z^2 \ge 12(xy+yz+zx) \implies x^2+2y^2+5z^2 \ge 2$$ Equality is when $x:y:z = 3:2:1$ and satisfying the constraint, i.e. when $x = \frac3{\sqrt{11}}, y = \frac2{\sqrt{11}}, z = \frac1{\sqrt{11}}$. In general for ...


2

$$|x^2-4|<2 \\ \implies -2<x^2-4<2 \\ \implies 2<x^2<6 \\ \implies \sqrt{2}<\pm x<\sqrt{6} \\ \implies \sqrt{2}<x<\sqrt{6},\ \ -\sqrt{2}>x>-\sqrt{6}$$


2

You want to show that $$2-\frac{1}{\sqrt{n}}+\frac1{(n+1)^4}\le 2-\frac1{\sqrt{n+1}}.\tag{1}$$ If (1) is true, then using induction hypothesis one gets $$1+\frac1{2^4}+\cdots+\frac{1}{n^4}+\frac{1}{(n+1)^4}\le 2-\frac1{\sqrt{n}}+\frac1{(n+1)^4}\le 2-\frac1{\sqrt{n+1}},$$ and we are done by induction. Showing (1) should be easy since ...


2

A symmetrical approach: $$\begin{align}x^2+2x+1\quad &=(x+1)^2&&=2(x^2+1)-(x^2+1-2x)\\ \\ x^2+1+\underbrace{2x}_{>0}\quad &=(x+1)^2&&=2(x^2+1)-\underbrace{(x-1)^2}_{\geq 0}\\ x^2+1\quad &<(x+1)^2&&\leq 2(x^2+1)\qquad \text{QED} \blacksquare \end{align}$$


2

As explained in the comments, the inequality $$\sum_kp_k(1-p_k)\lt\sum_kp_k\log p_k$$ cannot hold since the LHS is nonnegative and the RHS is nonpositive. To show that $$\sum_kp_k(1-p_k)\leqslant-\sum_kp_k\log p_k,$$ note that, for every positive $x$, $$1-x\leqslant-\log x.$$


2

Just squaring everything seems the best way: $$ \left(x+\frac1{2x}\right)^2 =x^2+1+\frac1{4x^2} $$ and $$ \begin{align} \left(\color{#C00000}{x+\frac1{2x}}-\frac1{8x^3}\right)^2 &=\color{#C00000}{x^2+1+\frac1{4x^2}}-\frac2{8x^3}\left(\color{#C00000}{x+\frac1{2x}}\right)+\frac1{64x^6}\\ &=x^2+1-\frac1{8x^4}+\frac1{64x^6} \end{align} $$ Now you need to ...


2

If $|x-2.5|<0.3$ then $2.2<x<2.8$. Therefore, $2<x<4$, or $|x-3|<1$.


2

Your proof is wrong. For one, arrows should be reversed. If $x<3$, then $-2x>-6$, so $10-2x>4$.



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