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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 Letf(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+pcancels 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 ...

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