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18

$$ \int_{0}^{1} x^2 (1-x)^2 e^{-x}\,dx = 14-\frac{38}{e},$$ but the LHS is the integral of a positive function on $(0,1)$.


7

We can see that $$f(x)=\frac{\ln x}{x}$$ is strictly increasing in $(0,e)$ and striclty decreasing in $(e,+\infty)$. Thus the equality $f(x)=f(y)$ cannot hold for the case of $x,y \in (0,e)$ or $x,y \in (0,+\infty)$. Since $x\lt y$, it follows that $x\in (0,e)$ and $y \in (e,+\infty)$. So let $y=rx$, for $r\in (1,+\infty)$ and we obtain $$(xr)^x = ...


6

$$\frac{\color{blue}{a+c}}{a+b}+\frac{\color{red}{b+d}}{b+c}+\frac{\color{blue}{c+a}}{c+d}+\frac{\color{red}{d+b}}{d+a}=\frac{\color{blue}{a+c}}{a+b}+\frac{\color{blue}{a+c}}{c+d}+\frac{\color{red}{b+d}}{b+c}+\frac{\color{red}{b+d}}{d+a}$$ ...


5

Consider the function $f(x)=x^{-1/x}$ with derivative $f'(x) = f(x)\cdot \frac{\log x - 1}{x^2}$. You want to show that $$ f(n+1) - f(n) \le \frac{1}{n+1} $$ for all integer $n$. For $x \ge 1$, clearly $f(x) \le 1$. Then $$ f(n+1) - f(n) = \int_n^{n+1} f'(x) dx \le \int_n^{n+1}\frac{\log x - 1}{x^2} dx $$ The integrand is decreasing for $x > e^{3/2} ...


5

$(n+1)!=2\cdot 3\cdot \dots\cdot(n+1)$ here a product of $n$ numbers all are at least 2 so the result follows...


4

HINT: $$\frac{(n+1)!}{2^n}=\frac{2}{2}\frac{3}{2}\frac{4}{2}\cdots \frac{n-1}{2}\frac{n}{2}\frac{n+1}{2}$$


4

Note that the integral $$ \int_{1}^{n} \log x\, dx = n \log n - n + 1 $$ can be bounded above by the sum $$ \sum_{k=2}^{n} \log k = \log(n!) $$ so, taking logarithms, it suffices to prove that $ 2 n \log n - 2n + 2 > n \log n $ for $ n > 2 $. We then have to show $ n \log n > 2n - 2 $. Define $ f(x) = x \log x - 2x + 2 $ and take derivatives to ...


4

For your induction argument, $$ k^k\geq(k+1)^{k-1}\iff\left(\frac{k}{k+1}\right)^k\geq\frac{1}{k+1}\iff\left(1-\frac{1}{k+1}\right)^k\geq\frac{1}{k+1}. $$ But the last inequality above holds because of Bernoulli's inequality: $$ \left(1-\frac{1}{k+1}\right)^k\geq 1-\frac{k}{k+1}=\frac{1}{k+1}. $$


4

Carl Friedrich Gauss would have solved this aged 7. $$\begin{align} n!^2 &= (1 \times 2 \times 3 \dots \times n) \times (1 \times 2 \times 3 \dots \times n) \\ &= (1 \times n) \times (2 \times (n-1)) \times (3 \times (n-2))\times \dots \times (n \times 1) \end{align}$$ The first and last of the $n$ factors are $n$, the others are all greater than ...


4

Here is a full proof. Let us start the discussion for general $n$. Denote $S = \sum_{i=1}^n a_i$. Since by AM-GM, $S \geq n \sqrt[n]{a_1a_2...a_n}$, we have $$1+\frac{n(n-2)\sqrt[n]{a_1a_2...a_n}}{2S} \geq \frac{S}{S - (n-2)\sqrt[n]{a_1a_2...a_n}}$$ Hence a tighter claim is (simultaneously defining $L$ and $R$): $$L = ...


4

If $a,b,c> 0$ and $abc=1$, then by Cauchy-Schwarz: $$\sum_{\text{cyc}}\frac{a}{2+bc}=\sum_{\text{cyc}}\frac{a^2}{2a+abc}=\sum_{\text{cyc}}\frac{a^2}{2a+1}\ge \frac{(a+b+c)^2}{2(a+b+c)+3}$$ And by AM-GM $a+b+c\ge 3\sqrt[3]{abc}=3$, so: $$(a+b+c)^2-2(a+b+c)-3=((a+b+c)-3)((a+b+c)+1)\ge 0$$ Equality holds if and only if $a=b=c=1$.


4

Think of a equilateral triangle $\triangle ABC$ where each side is $k$, then let $$\overline {AD}=A, \overline {DB}=a, \overline {BE}=B, \overline {EC}=b, \overline {CF}=C, \overline {FA}=c$$Where $D,E,F$ are points on $\overline {AB}, \overline {BC}, \overline {CA}$ respectively. Now, notice $$\frac{\sqrt ...


4

After your manipulation, the only reasonable claim is that: $$(a+b+c)^2 \geq a^2+b^2+c^2+6 $$ that is equivalent to: $$ ab+ac+bc \geq 3 $$ that follows from the AM-GM inequality: $$ ab+ac+bc = 3\cdot AM(ab,ac,bc) \geq 3\cdot GM(ab,ac,bc) = 3.$$ Equality occurs at $a=b=c=1$, as expected.


4

Without evaluating this integral: Note $1\le 2-\sin x\le 2$ on $\;[0,\frac\pi2]$, hence $\;\dfrac{x^2}2\le \dfrac{x^2}{2-\sin x}\le x^2 $, and by the positivity of the integral: $$\frac{\pi^3}{48}=\int_0^\tfrac\pi2\dfrac{x^2}2\,\mathrm d\mkern1mu x\le \int_0^\tfrac\pi2\dfrac{x^2}{2-\sin x}\,\mathrm d\mkern1mu x\le \int_0^\tfrac\pi2 x^2\,\mathrm d\mkern1mu ...


4

Here's an outline of how I got to this, rather than a pretty, cleaned up proof. First I worked out that $$ (-\log{f})'' = \frac{f'^2}{f^2}-\frac{f''}{f}, $$ which I thought might be useful because if you divide by $f^2 \geqslant 0$, you find terms that look like $f'/f$. Therefore the inequality is equivalent to showing $$ n (-\log{f})'' \geqslant ...


4

Let $a_1 = b_1 = 1$, $a_2 = b_2 = -1$, and let all the others be zero. Does there exist a constant $C$ for which $$|z + z^2| \le C |z - z^2|?$$ Hint: consider $z=1$.


3

By C-S $\sum\limits_{cyc}\frac{a}{2b^3+c}=\sum\limits_{cyc}\frac{a^2(a+c)^2}{a(a+c)^2(2b^3+c)}\geq\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}a(a+c)^2(2b^3+c)}$. Hence, it remains to prove that $(a+b+c)^2\left(\sum\limits_{cyc}(a^2+ab)\right)^2\geq9\sum\limits_{cyc}a(a+c)^2(2b^3+c)$, which is ...


3

Here's a proof. The formulae for the lenghts of the angular bisectors and the heights are known. (see e.g. https://en.wikipedia.org/wiki/Triangle ). Let $T$ be the area of the triangle. We have $$ l_a = \sqrt{bc (1 - \frac{a^2}{(b+c)^2})}$$ and $$h_a = \frac{2 T}{a}$$ and cyclic shifts of those. With these formulae, the required inequality gets $$ ...


3

The function $$f(x) = \log\left(\frac{x}{1-x}\right)$$ is concave on $(0,\tfrac12]$ (since $f''\leq 0$ on that interval) so by Jensen's inequality $$\frac{f(a_1)+\ldots+f(a_n)}{n}\leq f\left(\frac{a_1+\ldots+a_n}n\right).$$ Your inequality is an immediate consequence of this.


3

Since $2-\sin x\in[1,2]$ for $x\in [0,\frac\pi2]$, we have $$ \frac12\int_0^{\pi/2}x^2\,dx \le \int_0^{\pi/2}\frac{x^2}{2-\sin x}\,dx \le \int_0^{\pi/2}x^2\,dx. $$ Computing the integral of $x^2$ gives you the desired inequalities.


3

The first few convergents of the continued fraction representation of $e$ are $$ 2, 3, \frac{8}{3}, \frac{11}{4}, \frac{19}{7}, \frac{87}{32}, \frac{106}{39} $$ Since these convergents oscillate monotonically towards $e$, the last one works to prove that $e>\frac{19}{7}$. (If you know that $e$ is irrational, you can stop as soon as you get ...


3

Define $$f(x)=\sqrt x-\log(x+1)\implies f'(x)=\frac1{2\sqrt x}-\frac1{x+1}=\frac{x-2\sqrt x+1}{2\sqrt x(x+1)}=$$ $$=\frac{(\sqrt x-1)^2}{2\sqrt x(x+1)}\ge 0$$ since $\;x>0\;$ , and thus the function is monotonic ascendent, which means $$\forall\;x>0\;,\;\;\;f(x)\ge f(0)=0$$


3

Here is another way. You want to find maximum of $a(k-b)+b(k-c)+c(k-a)$. Note that for any $k$, this is linear in $a, b, c \in [0, k]$, so its extreme values will always be when $a, b, c \in \{0, k\}$. Take WLOG $a \ge b \ge c$ and this leaves only the trivial cases $(a, b, c) \in \{(0, 0, 0), (k, 0, 0), (k, k, 0), (k, k, k)\}$ to check.


3

Let $\lfloor \sqrt{n}\rfloor=k$. Note that with $k \le \sqrt{n} <k+1$, squaring each side would give that $$k^2 \le n < k^2+2k+1 \Leftrightarrow k^2 \le {n} \le k^2+2k $$ Since $\lfloor \sqrt{n}\rfloor=k$, We have that $$n+1 \le k^2+2k+1=(\lfloor \sqrt{n}\rfloor+1)^2$$


2

A non induction proof. It's enough to show that $$ 1<\frac{1^2}{n}\times\frac{2^2}{n}\times\cdots\times\frac{n^2}{n}. $$ First, suppose $n=2k$, $k\geq 1$. Then it's enough to show $\frac{j(n+1-j)}{n}\geq 1$ for all $j=1,\ldots,k\;$ and the inequality is strict for some $j$. Consider $$ j(n+1-j)-n=j(2k+1-j)-2k=(j-1)(2k-j) $$ which is $\geq 0$ for all $j$ ...


2

The rearrangement inequality implies that $$3(x^3+y^3+z^3)\ge (x+y+z)(x^2+y^2+z^2).$$ So the minimum is $\frac13$, attained when $x=y=z$. Since obviously $x,y,z\in(0,1)$, $$\frac{x^3+y^3+z^3}{x+y+z}<1\tag1.$$ On the other hand, if we set $y=z=\varepsilon$ and $x=\sqrt{1-2\varepsilon^2}$, then $$\lim_{\varepsilon\to 0}\frac{x^3+y^3+z^3}{x+y+z}=1.$$ That ...


2

Let $g(x)=f(x)-x$ and $m=\inf_{x\in[0,1]} g(x)$ then since $g$ is continuous on a compact so by Weierstrass theorem this minimum is attained i.e. there is $x_0\in[0,1]$ such that $g(x)\ge g(x_0)=m>0$. Take $\epsilon=\frac m2$.


2

Let $p,q,r \in \mathbb{R}_{>0}$ such that $(p^3,q^3,r^3)=(a,b,c)$. Then $(pqr)^3 = 1$ and hence $pqr = 1$. Thus $\sum_{cyc} \frac{a}{2+bc} \ge 1$ iff $\sum_{cyc} \frac{p^3 (pqr)}{2 (pqr)^2 + q^3 r^3} \ge 1$ iff $\sum_{cyc} \frac{p^4}{2 p^2 q r + q^2 r^2} \ge 1$. This technique is called homogenization, and many homogenous inequalities can be solved by ...


2

Note that $$\sum_{cyc}\frac{a}{bc+2}=\sum_{cyc}\frac{a^2}{2a+abc}=\sum_{cyc}\frac{a^2}{2a+1} \ge \frac{(a+b+c)^2}{2a+2b+2c+3} $$ Now one can use that $$\sum_{cyc}a \ge 3\sqrt[3]{abc}=3$$This establishes that $$(a+b+c-1)^2 \ge 4 \Leftrightarrow (a+b+c)^2 \ge 2(a+b+c)+3$$Thus, our proof is done, with equality at $a=b=c$.



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