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10

Since $(2n)!\geq 2^n n!$ for $n\geq0$ we have $$ \frac{e^x+e^{-x}}{2}=\sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}\leq \sum_{n=0}^\infty \frac{x^{2n}}{2^n n!}=e^{x^2/2}. $$


10

It's only true if $a(x)$ and $b(x)$ attain their minimum for the same value of $x$. For instance, consider $a(x) = (x+1)^2+1$ and $b(x) = (x-1)^2+1$. Clearly, $\min a = \min b = 1$. But $a(x)b(x) = x^4+4$, for which $\min ab = 4$. EDIT: The OP recently added the following to the question: From the below answers, i need more constraint to ...


8

This is the standard technique for this. This one is a bit tricky, but anyways. Let WLOG $x\le y\le z$ and let $y=a+x,z=b+x$. Now plug this into $$(x+y+z)\frac{1}{2}((x-y)^2+(y-z)^2+(z-x)^2)\ge C|(x-y)(y-z)(z-x)|$$ $$\iff(3x+a+b)(a^2-ab+b^2)\ge Cab(a-b)\quad \text{since } a\ge b$$ Now this is linear in $x$. Hence we may take $x=0$ to achieve the minimum of ...


7

Let $a_n=\frac{(n!)^2}{n^n}$, then $$r_n=\frac{a_{n+1}}{a_n}=(n+1)\big(\frac{n}{n+1}\big)^n\geq(n+1)e^{-1}$$ Since $n^n \le (n!)^2$ for $n=1,2$ and for $n\geq 2$, $r_n>1$ so we have the desired inequality for all $n$ by induction.


7

Hint: $$\dfrac{1}{11+a^2}\le\dfrac{7-a}{72}$$ $$\Longleftrightarrow (a-1)^2(a-5)\le 0$$ so $$\sum_{cyc}\dfrac{1}{11+a^2}\le \sum_{cyc}\dfrac{7-a}{72}=\dfrac{1}{3}$$


7

We make a couple of observations: If the constraints of the form $x_ix_j \leq 4^{-|i-j|}$ always remain in the form of inequalities, then we can always increase some of the entries without violating the conditions while increasing the sum. In particular, suppose there is an index i such that $x_ix_j \lt 4^{-|i-j|}$ for all j. Then we can keep increasing ...


6

Hint: $$\sqrt{\left(f(b)-f(a)\right)^2+\left(b-a\right)^2} = (b-a)\sqrt{\left(\frac{f(b)-f(a)}{b-a}\right)^2+1} $$


5

Take a look at the first few cases to get an idea what should happen. The first claim is $\frac11+\frac12\geq\frac12$, the second one is $\frac12+\frac13+\frac14\geq\frac12$, the third one is $\frac13+\frac14+\frac15+\frac16\geq\frac12$. So to get from $\frac1n+\dots+\frac1{2n}$ to $\frac1{n+1}+\dots+\frac1{2(n+1)}$ you need to subtract $\frac1n$ and add ...


5

Hint: We only need to prove this is true for $n=2$. Since if it's ok for $n=2$, then \begin{align} \frac{ab}{a+b} &= \frac{(a_1 + \sum_{i=2}^na_i)(b_1 + \sum_{i=2}^nb_i) }{a_1 + \sum_{i=2}^na_i + b_1 + \sum_{i=2}^nb_i} \\ &\geq \frac{a_1b_1}{a_1 + b_1} + \frac{(\sum_{i=2}^na_i)(\sum_{i=2}^nb_i)}{\sum_{i=2}^na_i + \sum_{i=2}^nb_i} \\ & \geq ...


5

Since you know that $x\neq 0$ you can multiply by $x$. Just remember that $x<0$. So, multiplying by $x$ reverses the inequalities. Then you get $-5x>1>0$. This gives $-5x>1$. Now divide by $-5$.


5

You don't have to necessarily put $x$ to the center directly. One way to solve such double inequality is to split it to a system of two inequalities: $$\begin{cases}-5<\frac1x\\ \frac1x<0\end{cases}$$ From the second one you get $x<0$, and from the first one you can obtain $x<-\frac15$ (not forgetting that $x<0$). Now the answer is ...


5

In this case you want to have quadratic terms in the RHS, so use Cauchy in the following form: $$\left(\sum_{cyc} \frac{a^2}b \right) \left(\sum_{cyc} a^2b \right) \ge \left(a^2+b^2+c^2\right)^2 \tag{1}$$ Then it is enough to show that $$a^2+b^2+c^2 \ge 3\sum_{cyc} a^2b \tag{2}$$ Homogenising, $$\iff \left( a^2+b^2+c^2\right) (a+b+c) \ge 3\sum_{cyc} a^2b ...


4

$$1-\frac{x^2}{n}\overset{?}{\leq}\left(1-\frac{x^2}{n^2}\right)^n$$ is a good starting point. You can assume $|x|<\sqrt{n}$, since otherwise the inequality is trivial, with the LHS being non positive and the RHS being positive. Consider the logarithm of both sides. Then: $$\log\left(1-\frac{x^2}{n}\right)\leq n\log\left(1-\frac{x^2}{n^2}\right)$$ is a ...


4

Hint: $$\left(x^{n-m}\right)^{\frac1{n-m}}\cdot\left(\frac a{x^m}\right)^{\frac1m}=a^{\frac1m}$$ is certainly constant.


4

The next thing you can try is Cauchy again $$\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)(b+c+a)\geq (a+b+c)^2.$$ So $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c.$$ Now you can eliminate $a+b+c$ from your first inequality.


4

Apply Cauchy–Schwarz on the vectors $u=(f'(x),1)$ and $v=(f(b)-f(a),b-a)$ to find $$f'(x)(f(b)-f(a)) + (b-a) \leq \sqrt{f'(x)^2 + 1} \sqrt{\left(f(b)-f(a)\right)^2 + \left(b-a\right)^2}$$ Now integrate the inequality over $[a,b]$ to obtain the result $$\sqrt{\left(f(b)-f(a)\right)^2 + \left(b-a\right)^2} \leq \int_a^b\sqrt{f'(x)^2 + 1}dx$$ ${\bf ...


4

$$ n \ge r \implies n(r-1) \ge r(r-1) \implies nr - n - r(r-1) \ge 0 \implies r(n-r + 1) \ge n $$


4

I'm going to reformulate the problem in a slightly more convenient way (I rename $k_i-2$ as $k_i$). Consider a sequence $\{(x_i,k_i)\}_{i=1}^n\subset \mathbb{R}^+\times \mathbb{Z}$ such that $$ x_ik_i=(x_{i+1}-x_i)-(x_{i}-x_{i-1}). $$ Prove that $0\leq \sum_{i=1}^nk_i<n.$ Upper Bound: Since $x_i(k_i+2)=x_{i+1}+x_{i-1}>0$, each $k_i\geq -1$. ...


3

HINT: Use the fact that $$ r (n+1-r) \geq n $$ (quadratic polynomial in $r$ with equality at $r=1$ and $r=n$) when $1\leq r\leq n$. Then, take the product over $r$: $$ \left(n! \right)^2=\prod_{r=1}^n r(n+1-r) \geq \prod_{r=1}^n n=n^n $$


3

The proof is nearly correct, but induction is unnecesary: $$\frac{1}{n} + \frac{1}{n+1} +\ldots + \frac{1}{2n} \geq \underbrace{\frac{1}{2n} + \frac{1}{2n}+\ldots +\frac{1}{2n} \frac{1}{2n}}_{n\text{ times}} = n\frac{1}{2n} = \frac{1}{2}$$


3

By scaling, which does not change the ratio of maximum to minimum, we can assume $$ \int_a^bf(x)^2\,\mathrm{d}x=\int_a^bg(x)^2\,\mathrm{d}x=1\tag{1} $$ Trivially, we have $$ \int_a^bf(x)g(x)\frac{f(x)}{g(x)}\,\mathrm{d}x=\int_a^bf(x)^2\,\mathrm{d}x=1\tag{2} $$ and $$ \int_a^bf(x)g(x)\frac{g(x)}{f(x)}\,\mathrm{d}x=\int_a^bg(x)^2\,\mathrm{d}x=1\tag{3} $$ Since ...


3

As discussed on the comments, as the terms only make sense for $n \ge 2$, we can just compare the two sides term by term. $1 - x_i \le 1 \le n - 1$, so $\frac{x_i}{\sqrt{1-x_i}} \ge \frac{x_i}{\sqrt{n-1}}$ Now just sum each side for i going from 1 to n.


3

Here is a proof of a reverse Hölder inequality proven in a manner very similar to the proof of the reverse Cauchy-Schwarz inequality in my other answer. In what follows, $p,q\gt1$ and $\frac1p+\frac1q=1$. The result requested in the question is simply the case $p=q=2$. By scaling, which does not change the ratio of maximum to minimum, we can assume $$ ...


3

We use continuity of $f$ and $g$, to prove that the integral inequality: $$\displaystyle \int_a^b f^2\int_a^bg^2 \le \dfrac{1}{4}\left(\sqrt{\dfrac{M_1M_2}{m_1m_2}}+\sqrt{\dfrac{m_1m_2}{M_1M_2}}\int_a^bfg\right)^2$$ follows from proving the discrete case of the inequality, taking $a_k = f(a+\frac{b-a}{n}k)$ and $b_k = g(a+\frac{b-a}{n}k)$, for ...


3

The rank does not behave well under sum Example $0=A+(-A)$ for any $A$. On the other hand, the rank is subadditive: $rank(A+B)\leq rank(A)+rank(B)$ Proof. I denote by "span of a set" the vector space generated by that set. The rank of a matrix is the dimension of the span of the set of its columns. The span of the columns of $A+B$ is contained in the ...


3

Set $y = \dfrac{|x|-|1-x|}{|x|}= 1 - \lvert \frac1x-1 \rvert$. Note $y > 1 $ is not possible, and $y \in [0, 1]$ when $x \ge \frac12$. Otherwise $y$ is negative, (so what?). So the only case to consider is $y \in [0, 1]$, which means $2x-1< 8-x \implies ...$


3

Since the inequality is homogeneous, WLOG assume that $a+b+c = 1$. Then, the inequality becomes $\dfrac{a}{(1-a)^2}+\dfrac{b}{(1-b)^2}+\dfrac{c}{(1-c)^2} \ge \dfrac{9}{4}$. Since the function $f(x) = \dfrac{x}{(1-x)^2}$ is concave up for $x > 0$, by Jensen's Inequality, we have: $f(a)+f(b)+f(c) \ge 3f\left(\dfrac{a+b+c}{3}\right) = ...


3

Let $x=b+c,y=c+a,z=a+b$, then $a=\frac{y+z-x}{2},\dots$. Your inequality becomes $$(x+y+z)\left(\frac{y+z-x}{x^2}+\cdots\right)\geq 9.$$ Write $y+z-x=(x+y+z)-2x,\dots$ we need to show $$(x+y+z)^2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right) -2(x+y+z)\left(\frac1x+\frac1y+\frac1z\right)\geq 9.$$ Use ...


3

Here is a sketch of what you might be looking for: Showing $\tan x > x$ is equivalent to showing $\sin x - x \cos x > 0$, since $\cos x > 0$ on $(0,\pi/2$). The series for $\sin x - x \cos x$ is $\displaystyle\sum_{j=1}^{\infty} \dfrac{(2j)x^{2j+1}}{(2j+1)!} = x^3/3 - x^5/30 + x^7/840 - x^9/45360 \ldots$ Group the terms in pairs: $(x^3/3 - ...


3

If $x > 1 \quad \Rightarrow \quad x^2 > x \quad \Rightarrow x^3 > x^2 \ \ldots$. By induction, $x^{n+1} > x^n$. Thus, $x^m \geq x^{n+1} > x^n$.



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