Tag Info

Hot answers tagged

6

For any $y \ge 0$, notice $$e^y - 1 = \int_0^y e^x dx \ge \int_0^y (1+x) dx \ge \int_0^y \left(1+\frac{x} {\sqrt{1+x^2}}\right)dx = y + \sqrt{1+y^2} - 1$$ we have this little inequality: $$\sqrt{1+y^2} - y = \frac{1}{\sqrt{1+y^2} + y} \ge e^{-y}$$ Using MVT, we can find a $\xi \in (y,\sqrt{1+y^2})$ such that $$\sinh\sqrt{1+y^2} - \sinh(y) = ...


6

Hint $$\sqrt{2}x-\sqrt{x^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{x},x>0$$ it is easy to prove by derivative. so $$\sqrt{2}a-\sqrt{a^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{a}\tag{1}$$ $$\sqrt{2}b-\sqrt{b^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{b}\tag{2}$$ $$\sqrt{2}c-\sqrt{c^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{c}\tag{3}$$ $(1)+(2)+(3)$ ...


5

Here's a solution with very little calculus. First, an identity: \begin{align*} \sinh^2(a+b) - \sinh^2(a-b) &= (\sinh a\cosh b + \cosh a\sinh b)^2 - (\sinh a\cosh b - \cosh a\sinh b)^2 \\ &= 4\sinh a \cosh a\sinh b \cosh b \\ &= \sinh(2a)\sinh(2b) \end{align*} Taking $a=e^x/2$ and $b=e^{-x}/2$, we get \begin{align*} \sinh^2(\cosh x) - ...


5

The Arithmetic Mean-Geometric Mean Inequality states that for any non-negative numbers $a$ and $b$, we have $AM := \dfrac{a+b}{2} \ge \sqrt{ab} =: GM$. This can be visualized as follows: Now, set $a = z$ and $b = \dfrac{1}{z}$ to visualize what you wanted.


5

The AM-GM inequality states that $x+y+z\geq 3(xyz)^\frac{1}{3}$ and is an extremely useful inequality to know. By AM-GM inequality, we have $ab+bc+ca\geq 3(abc)^\frac{2}{3}$. However, since $abc=1$, the inequality follows.


5

Let $$f(x)=\frac{\sqrt[3]{x+1}}{x}$$ Take logarithms: $$g(x)=\ln f(x)=\frac13\ln(x+1)-\ln x$$ Now, $$g'(x)=\frac1{3(x+1)}-\frac1x$$ which is negative for $x\ge 1$. That is, $f$ is decreasing.


5

hint: Use $|A| > |B| \iff A^2 > B^2 \iff (A-B)(A+B) > 0$. Apply this property to $A = x^2-2x-3, B = x^2+7x-13$


5

A first observation is that $|x+y|\geqslant|x-y|$ if and only if $xy\geqslant 0$, so defining $$f(x,y):=|x+y|-|x-y|,$$ the positivity of $f$ is linked to that of $xy$. We would like to find a more tractable expression for $f$. Assume that $x\gt 0$ and $ y\gt 0$. Then $f(x,y)=x+y-|x-y|=2\min\{x,y\} =\min\{|x|,|y|\}$. Since $f(-x,-y)=f(x,y)$ we get ...


4

Note that $$a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\left((a-b)^2+(b-c)^2+(c-a)^2\right).$$


4

Just to be clear on what it would take to decide this based on rational approximation: we want to compare the ratio $$ f(x)=\frac{x^x}{(x-1)^{x+1}} $$ to unity at $x=(\pi + 1)$. Its logarithm is $$ \begin{eqnarray} g(x)=x\log x - (x+1)\log(x-1) &=& x\log x - (x+1)\log x - (x+1)\log(1-1/x) \\ &=& -\log x - (x+1)\log(1-1/x) \\ &=& ...


3

Hint We have $x^2+y^2 \geq 2 x y .$ Write down the three inequialities for $a, b$ and $a,c$ and $b,c$ and then add them.


3

multiply by $2$ ,then you have : $$2a^2+2b^2+2c^2 ≥ 2ab+2bc+2ac\\ 2a^2+2b^2+2c^2 -( 2ab+2bc+2ac) ≥0\\(a^2+b^2-2ab)+(b^2+c^2-2bc)+(a^2+c^2-2ac)≥ 0\\$$


3

Hint: Let $f(x)=\dfrac{4x^2-1}{e^x}$. Note that this function decreases for $x \ge 3$. Further, we can check that $f(2)> \int_2^3 f(x) dx$. $$\implies \int_{2}^3 f(x)dx +\int_{3}^\infty f(x)dx < f(2)+\sum_{3}^\infty f(n) < f(2)+\int_{2}^\infty f(x)dx $$ $$\implies \int_{2}^\infty f(x)dx < \sum_{2}^\infty f(n) < f(2)+\int_{2}^\infty f(x)dx ...


3

You can use the harmonic-geometric inequality : the harmonic means of $a,b,c$ is defined by$$\frac1H=\frac 13\Bigl(\frac1a+\frac1b+\frac1c\Bigr)$$ and it is known that $H\le G$. Since $G=1$, $$\frac1H=\frac{ab+bc+ca}{3}\ge \frac1G=1.$$


3

Plot $xy = 1$ and $x+y = 2$. Interpret $x=z$ and $y=1/z$ (which can only hold on the blue hyperbola). Or vice versa. The fact the plot is symmetric under swapping $x$ and $y$ is important! (i.e. symmetric under reflecting across the line $x=y$) It may help to see some more lines depicting how $x+y$ varies: The red line $x+y=2$ is the smallest line ...


3

For $x\ge0$, the Mean Value Theorem says that for some $\sinh(x)\lt\xi\lt\cosh(x)$, $$ \begin{align} \sinh(\cosh(x))-\sinh(\sinh(x)) &=\cosh(\xi)(\cosh(x)-\sinh(x))\\ &\gt\cosh(\sinh(x))\,e^{-x}\tag{1} \end{align} $$ Furthermore, $$ \cosh(\sinh(x))-\sinh(\sinh(x))=e^{-\sinh(x)}\tag{2} $$ Therefore, subtracting $(2)$ from $(1)$, then applying ...


3

Let $t=\sinh x$. Now we can square the inequality and instead try proving $$\sinh^2(\sqrt{1+t^2})=\sinh^2(\cosh x) \ge \cosh^2(\sinh x)=1+\sinh^2t$$ So it is enough to show $f(t) = \sinh^2(\sqrt{1+t^2})-\sinh^2t-1 \ge 0$. As $f$ is even and $f(0)> 0$, it is enough to show it is increasing for positive $t$. Hence we look at $$f'(t) = \frac{t\sinh ...


2

The inequality does not always hold. Put $X=A^{1/2}$ and $Y=B^{1/2}$. The inequality is equivalent to $|||X^2Y^2|||\le|||XY|||^2$ for every pair of positive definite matrices $X,Y$ and for every unitarily invariant matrix norm $|||\cdot|||$. Now, take $X=\pmatrix{2&0\\ 0&1},\ Y=\pmatrix{2&1\\ 1&1}$ and the operator norm (induced 2-norm), we ...


2

Well, what you wrote is clearly not true, it is only true for $x\in (0,\infty)$. You could just try plotting the function $x\mapsto x+\frac1x$ and see that it is always above $2$. http://www.wolframalpha.com/input/?i=plot+z+%2B+1%2Fz Or, you can plot $x\mapsto x$ and $x\mapsto \frac1x$ and try to understand what is happening on $(0,\infty)$. It's clear ...


2

HINT: note that $$10^x=2^x\cdot 5^x$$


2

Let $y = [x] \Rightarrow x-1 <y \leq x\Rightarrow x < y+1 \leq x+1\rightarrow y+1 = [x+1]\Rightarrow [x]+1=[x+1]$


2

Setting $F(x)=\int_{-R}^xf(y)\mathrm dy=H\star f(x)$ and $G(x)=\int_{-R}^xyf(y)\mathrm dy$, we have $F'(x)=f(x)$, $G'(x)=xf(x)$, $F(-R)=F(R)=0$. Proceed with integration $$\begin{split}A&=\iint f(x)f(y)|x-y|\,\mathrm dy\\ &=\int_{-R}^Rf(x)\mathrm dx\left[x\int_{-R}^xf(y)\mathrm dy-\int_{-R}^xyf(y)\mathrm dy-x\int_x^Rf(y)\mathrm ...


2

This is basically the same argument as the probability approach used in the original solution, but somewhat simplified and hopefully easier to understand. To help avoid special cases for the $n_i$, let's start by appending zeroes to the list $a_1,\ldots,a_N,0,\ldots,0$ so it gets length $2^k-1\ge N$. We will make a solution to this consisting of ...


2

Assume that $k^2>k+1$. Then $(k+1)^2=k^2+2k+1>k+1+2k+1=3k+2>k+2,$ because $k>0$.


2

here is another way which is not use derivative: first we need to know: $\sqrt{\dfrac{x+y+z}{3}} \ge \dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}$ (1) to prove it, we have $\sqrt{\dfrac{x+y}{2}} \ge \dfrac{\sqrt{x}+\sqrt{y}}{2} \implies \sqrt{\dfrac{x+y+z+t}{4}} \ge \dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{t}}{4}$ let $t= \dfrac{x+y+z}{3}$ we have (1) squre ...


2

Consider the function $$f(x)=x\log(x)+px - q$$ The first derivative $$f'(x)=\log (x)+1+p$$ cancels at a single place $x=e^{-(p+1)}$ and at this point the value of the function is $-(q+e^{-(p+1)})$. The second derivative being always positive, then this point corresponds to a minimum. So, the function (which is defined for $x \geq 0$) starts from $-q$, goes ...


2

As stated, part (c) is false (as Alex Ravsky shows). Suppose we change the definitions to: \begin{align} &M = \sup_{t\in [0,2]} f'''(t)\\ &m = \inf_{t\in[0,2]} f'''(t) \end{align} Then part (c) can be proven from part (b) by defining $h(x) = f(x) - cx(x-1)(x-2)$ for a suitable value of $c$ such that $h'''(x) \in [-(M-m)/2, (M-m)/2]$ for all $x \in ...


2

Yes your proof is correct. Excellent work reducing the question about multiplying inequalities to a more familiar one of adding inequalities. The only thing I would mention is that taking logarithms and exponentiating are monotone increasing operations. If they were monotone decreasing, the inequalities would flip.


2

Let $a^2=3xy, b^2=yz, c^2=3zx$, then $a^2+b^2+c^2=3$ and we need to show $$\sum_{cyc} \frac{bc}{a(3+a)} \le \frac3{4abc} \iff \sum_{cyc} \frac{b^2c^2}{3+a} \le \frac34$$ Elementary methods seem difficult on that, though we can use the $uvw$ trick. Rationalising denominators and using $3u=a+b+c, \; 3v^2=ab+bc+ca \implies 2v^2=3u^2-1,\; w^3=abc$, you can ...


2

We know that $(\sqrt{a}-\sqrt{b})^2 \geq 0$ since squares are always positive. Expanding this gives $a-2\sqrt{ab}+b \geq 0$, or that $a+b\geq 2\sqrt{ab}$. Multiplying both sides by $\sqrt{ab}$ gives the inequality you are trying to prove. While it might be easier intuitively to think about it in this way, mathematicians usually go about proving things in ...



Only top voted, non community-wiki answers of a minimum length are eligible