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26

Sketch the graph of $x^2$ (it's a parabola opening upwards with vertex in $(0,0)$) and sketch the line $y=10$. They intersect in $x=-\sqrt{10}$ and $x=\sqrt{10}$, and the sketch immediately gives the solution to the inequality: $$x<-\sqrt{10} \vee x>\sqrt{10}$$


18

Using $a^2 - b^2 = (a+b)(a-b)$, we get $(x-\sqrt{10})(x+\sqrt{10}) > 0$, which mean $x+\sqrt{10}$ and $x-\sqrt{10}$ have the same sign


7

By AM-GM, we have $$\frac{\frac{1}{x}+\sqrt{x}+\sqrt{x}}{3}\ge\sqrt[3]{\frac{1}{x}\sqrt{x}\sqrt{x}}=1$$ Hence $$\frac{1}{x}+2\sqrt{x}\ge3$$ as desired.


6

$$2\sqrt{x}-\left(3-\frac1x\right)=\left(2\sqrt{x}+1\right)\cdot\left(1-\frac1{\sqrt{x}}\right)^2\geqslant0$$


6

You could also square everything $$ |f(x)| < |g(x)| \Leftrightarrow |f(x)|^2 < |g(x)|^2 \\ \Leftrightarrow f(x)^2 < g(x)^2 \\ \Leftrightarrow 0< g(x)^2-f(x)^2 \\ \Leftrightarrow 0< (g(x)-f(x))(g(x)+f(x)), \\ $$ which means that $g(x)-f(x)$ and $g(x)+f(x)$ have the same sign.


6

Sure, your proof is pretty short, but it is only short for someone who already did some analysis. For example, you used the following theorems (which is true, of course) in your proof: If $f$ is convex on $\mathbb R$, then for any tangent function $g$, we know that $g(x)<f(x)$ for all $x$. and If $f$ is twice differentiable on $\mathbb R$ and ...


5

I don't know how you have defined the exponential function, but you could use the fact that it has the following series expansion $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +\cdots$$


5

Another way to see it algebraicaly/analyticaly is this: $(-x)^2 = x^2 > 10$ then you have 2 conditions: a) $-x > \sqrt{10} \implies x < -\sqrt{10}$ b) $x > \sqrt{10}$ which both provide solutions


5

Notice that the first four fractions all cancel out. So now we have: $$ \left(\frac{3}{4}\right)\left(\frac{4}{5}\right) \cdots \left(\frac{k-1}{k}\right)\left(\frac{k}{k+1}\right) \geq \frac{1}{8} $$ But observe that all but two of the numbers in the fractions in the LHS cancel in pairs (the $4$'s cancel, the $5$'s cancel, ..., the $(k-1)$'s cancel, and the ...


4

Consider $m$ number of $\dfrac{\sin^2\alpha}m$ and $n$ number of $\dfrac{\cos^2\alpha}n$ As each of the term $\ge0$ for real $\alpha;$ using AM, GM inequality $$\frac{m\cdot\dfrac{\sin^2\alpha}m+n\cdot\dfrac{\cos^2\alpha}n}{m+n}\ge \left[\left(\dfrac{\sin^2\alpha}m\right)^m\left(\dfrac{\cos^2\alpha}n\right)^n\right]^{\dfrac1{m+n}}$$


4

By aM-GM: $$\frac{2a^3+b^3}3\ge(a^6b^3)^{\frac13}=a^2b$$ $$\frac{2b^3+c^3}3\ge(b^6c^3)^{\frac13}=b^2c$$ $$\frac{2c^3+a^3}3\ge(c^6a^3)^{\frac13}=c^2a$$ Add them.


4

Another (perhaps more systematic?) approach: $$x^2 > 10 \implies |x| > \sqrt{10} \implies x > +\sqrt{10}\ \lor\ x < -\sqrt{10}$$


4

The Arithmetic-harmonic mean inequality is actually the usual way of doing this. We have $ \frac{x_1+\cdots+x_n}{n}\geq \frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}} $ in general. Write it out for $n=3$, and stare at it for a bit - it's a very simple manipulation.


4

Since $x=\frac{1}{m}(1-y)$, we have: $$\left(1+\frac{1}{x}\right)\left(1+\frac{m}{y}\right)=\left(1+\frac{m}{1-y}\right)\left(1+\frac{m}{y}\right),\tag{1}$$ and since $f(x)=1+\frac{m}{x}$ is a log-convex function, the minimum of the RHS of $(1)$ over $(0,1)$ occurs when $y=\frac{1}{2}$, and the value of such a minimum is just $(1+2m)^2$. The answer is so ...


4

Suppose there exist $a, b, c, x, y, z > 0$ such that $$a+b+c=4,$$ $$ax+by+cz = xyz,$$ $$x+y+z \leq 4.$$ Observe that we have $$axyz + bxyz + cxyz = 4xyz,$$ so that $$axyz + bxyz + cxyz = 4ax + 4by + 4cz,$$ and hence $$ax(yz-4) + by(xz-4) + cz(xy-4) = 0.$$ Since $x+y+z \leq 4$ and $a, b, c, x, y, z > 0,$ what follow are the inequalities $$xy, yz, xz ...


4

Expand $(a^2b+b^2c+c^2a)^2$ and use the fact that $abc=1$ to reduce it and then factorise to get, $$ a^2(a^2b^2+2c) + b^2(b^2c^2+2a) + c^2(c^2a^2+2b) $$ Using the fact $abc=1$ again, $$ a^2\left(\frac{1}{c^2}+2c\right) + b^2\left(\frac{1}{a^2}+2a\right) + c^2\left(\frac{1}{b^2}+2b\right) $$ and then it's easy to show that $\frac{1}{x^2}+2x\geq 3$ for all ...


4

Since it seems the PDF linked by Macavity is not accessible to everyone, I reproduce it here :


3

Hint: If all three numbers are negative, then: $$ab > \left(\frac{a+b+c}{3}\right)^2 \hspace{1cm} ac > \left(\frac{a+b+c}{3}\right)^2 \hspace{1cm} bc > \left(\frac{a+b+c}{3}\right)^2 \hspace{1cm}$$ Therefore, if we multiply the three inequalities: $$a^2b^2c^2 > \left(\frac{a+b+c}{3}\right)^6$$ Or equivalently: $$\left(\sqrt[3]{abc}\right)^6 ...


3

$$\sum [(a+b+c)^2-9bc]=3\sum[a^2+b^2+c^2-ab-bc-ca]=\frac32\sum (a-b)^2\ge0$$ If each $(a+b+c)^2-3bc<0,$ $$\sum [(a+b+c)^2-3bc]<0$$


3

The more precise part makes perfect sense: sure, for every $n$ you can apply Markov's inequality to $X(n)$ and $f(n)$ (assuming $X(n)$ is nonnegative), obtaining $$\mathbb P(X(n)\ge f(n))\le \frac{\mathbb E(X(n))}{f(n)}$$ The first sentence of your question is meaningless. Applying some operation to every term in a sequence of numbers than tends to ...


3

We have that $\int_{a}^{b}\left(\lambda f(x)-g(x)\right)^2dx\ge0$ for all $\lambda$, so $\displaystyle\lambda^{2}\left(\int_{a}^{b}(f(x))^2dx\right)-2\lambda\left(\int_{a}^{b}f(x)g(x)dx\right)+\int_{a}^{b}(g(x))^2dx\ge0$ for all $\lambda$. Therefore ...


3

Let $a=y-x$ and $b=q-p$. We want $a\ge 0$ OR $b\ge 0$. Equivalently, we want $\max\{a,b\}\ge 0$. We have $\max\{a,b\}=\frac{a+b}{2}+\frac{|a-b|}{2}$. (This is a standard expression for the max of two numbers--geometrically the midpoint of the two numbers, plus half the distance between them.) So for the inequality to hold, we want ...


3

My favourite proof is letting $f(x)=e^x-x-1$. $f(0)=0$ and it's easy to show that it has a minimum at $x=0$. However you do it, you are somehow going to go through derivatives or calclulus, explicitly or implicitly. EDIT: A completely alternative proof: Using the taylor series ($e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +...$), it is easy to show the ...


3

When $x>=0$, $e^x>=1+x$. When $x <-1$, $\;1+x < 0$, so $e^x>=1+x$ When $ -1\leq x \lt 0$, then lets take $y = -x$, so, $$e^x=e^{-y}=(1-y)+y^2(1/2!-y/3!+\cdots>=(1-y)+y^2((1/2!-1/3!)+y^2(1/4!-1/5!)+\cdots)\geq (1-y)\geq(1+x)$$ [Remember, $(1/n!-1/(n+1)!>0$]


3

Writing it as $m^m>(m+1)^{m-1}$ for $m>1$ might make it a little easier. First, show $\binom{m-1}{i}\leq m^{m-1-i}$, with strict inequality when $0\leq i<m-1$. Then, for any positive $m$, $$(m+1)^{m-1}=\sum_{i=0}^{m-1}\binom{m-1}{i} m^i\leq \sum_{i=0}^{m-1} m^{m-1} = m^m$$ with strict inequality when $m>1$.


3

This is a well-known fact in the theory of Farey series. You have $aq>bp$ and $pd>qc$, so the integers $\delta_1=aq-bp$ and $\delta_2=dp-cq$ are both $\geq 1$. Now $$ d\delta_1+b\delta_2=(ad-bc)q=q $$ It follows that $q\geq b+d$.


3

In general, using the power mean inequality, $$\sqrt[n+1]{\frac{x^{n+1} + y^{n+1} + z^{n+1}}{3}} \ge \sqrt[n]{\frac{x^n + y^n + z^n}{3}}$$ Raising both sides to the same power, $$\frac{x^{n+1} + y^{n+1} + z^{n+1}}{3} \ge \frac{x^n + y^n + z^n}{3} \cdot \sqrt[n]{\frac{x^n + y^n + z^n}{3}}$$ But, again by the power mean inequality, $$\sqrt[n]{\frac{x^n + ...


3

Given that: $$\frac{x_{n+1}}{y_{n+1}}=\frac{x_n}{y_n}+\frac{1}{x_n y_n}$$ we have: $$\frac{x_{n+1}}{y_{n+1}}=\frac{1}{x_n y_n}+\frac{1}{x_{n-1} y_{n-1}}+\ldots+\frac{1}{x_1 y_1}+\frac{x_1}{y_1}$$ or just: $$\frac{x_{n+1}}{y_{n+1}}=\frac{1}{x_n\cdot\ldots\cdot x_1}+\frac{1}{x_{n-1}\cdot\ldots\cdot x_1}+\ldots+\frac{1}{x_1}+2.$$ Hence we need to prove that: ...


2

Do as you said and we only have to prove $$2\sum_{cyc}\sqrt{xy+x+y+1}\le xy+yz+xz+6-x-y-z$$ This is right, since $\displaystyle x+y+z=-xyz$ and $x,y,z,xy,yz,xz\in(-1;1)$, we have: $2\sqrt{xy+x+y+1} =2\sqrt{xy+x+y+1-x-y-z-xyz} =2\sqrt{xy-xyz+1-z} =2\sqrt{(1-z)(1+xy)} \le 1-z+1+xy$ Do this with the other two and sum up, we have the inequality.


2

From Chebyshev's sum inequality we have \begin{align*} \frac{x^{n+1}+y^{n+1}+z^{n+1}}{3}\geq \frac{x^n+y^n+z^n}{3}\cdot\frac{x+y+z}{3}. \end{align*} By AM-GM we have $\frac{x+y+z}{3}\geq\sqrt[3]{xyz}=1$ and that proves the desired inequality.



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