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16

Due to symmetry, assume wlog that $a\geqslant b\geqslant c\geqslant d\geqslant 0$. Since I will use this fact later, I am proving it before the cases: Lemma: We have that $2\geqslant a+d$. In fact, this follows from Cauchy-Schwarz: $$2= 2\sqrt{\frac{a^2+b^2+c^2+d^2}{3}}\geqslant 2\sqrt{\frac{a^2+3d^2}{3}}=\sqrt{1+\frac13}\cdot \sqrt{a^2+3d^2}\geqslant a+d... 15 Since AA^T = I, we have$$\det (A+B)= \det A(I+A^TB) = \det A \,\det(I+A^TB).$$Now, |\det A| = 1 since A is orthogonal, and A^TB is orthogonal as well so all its eigenvalues \lambda_1, \ldots, \lambda_n are on the unit sphere in \Bbb{C}. Therefore$$|\det(A+B)| = |\det(I+A^TB)| = |(1+\lambda_1)\cdots(1+\lambda_n)| \le (1+|\lambda_1|)\cdots(1+|\...

8

Mechanodroid's solution is great. Here's a slightly different way to look at it. Suppose $\lambda$ is an eigenvalue of $A+B$ with corresponding unit-length eigenvector $v$. Then, $$|\lambda| = \|\lambda v\| = \|(A+B)v\| = \|Av+Bv\| \le \|Av\|+\|Bv\| = \|v\|+\|v\| = 2\|v\| = 2,$$ where we have used the fact that $\|Av\| = \|v\|$ and $\|Bv\| = \|v\|$ since $A,... 7 Since$a^2+b^2+c^2+d^2 = 3$defines a compact set in$\mathbb{R}^4$,$f(a,b,c,d)=abcd+3-a-b-c-d$will have a global minimum and maximum over that set, that will occur (can be easily justified) in critical points of the Lagrangian $$L(a,b,c,d,\lambda) = abcd+3-a-b-c-d -\lambda(a^2+b^2+c^2+d^2-3)$$ So, just compute the critical points, compute the value of$...

6

A possible way is to set $x = \tan t$ with $t \in \left(0,\frac{\pi}{2}\right)$. So, the inequality is equivalent to $$\frac{\tan t}{1+\frac 2{\pi} \tan t}< t \text{ for } t \in \left(0,\frac{\pi}{2}\right)$$ or after rearranging $$\tan t < \frac t{1-\frac 2{\pi}t} \text{ for } t \in \left(0,\frac{\pi}{2}\right)$$ But this is true because for $t \in ... 6 Note that $$(x+y)^3 = x^3+y^3+3xy(x+y).$$ Using the AM-GM inequality, we have $$xy\sqrt[3]\frac{x^3+y^3}{2} = \frac{1}{x+y} \cdot \sqrt[3]{\frac12 \cdot \left[xy(x+y)\right]^3 \cdot (x^3+y^3)}$$ $$\leqslant \frac{1}{x+y} \cdot \sqrt[3]{\frac{1}{2}\left(\frac{3 xy(x+y) + x^3+y^3}{4}\right)^4} = \frac{(x+y)^3}{8}.$$ 5 Fix$n\in\Bbb N$. The claim is trivial in the case$k=1$, so by induction assume the claim is true for some$k<n. Then we get: \begin{align*} \left(1+\frac{1}{n}\right)^{k+1}&<\left(1+\frac{k}{n}+\frac{k^2}{n^2}\right)\left(1+\frac{1}{n}\right)\\ &=1+\frac{k}{n}+\frac{k^2}{n^2}+\frac{1}{n}+\frac{k}{n^2}+\frac{k^2}{n^3}\\ &\overset{k<n}{&... 5 It is true that this falls under the realm of probability theory, because ifX$is a real valued random variable with density$f(x)$i.e.$P(X \in (a,b)) = \int_a^b f(x)dx$for all$a<b$, then :$\int_{-\infty}^{\infty} f(x)dx = 1$and$f(x) \geq 0$assert that$f$is a well-defined density and$X$a well defined random variable.$\int_{-\infty}^\infty ...

4

lower bound: Notice that $$\cos ^2 \theta_1=\sin^2 \theta_2+\sin^2 \theta_3+\cdots+\sin^2 \theta_{10}$$ similarly $$\cos^2 \theta_i=\sum_{j=1 \rightarrow 10 , j\neq i }\sin^2 \theta_j$$ $$S_i=\sum_{j=1\rightarrow 10,j\neq i} \sin \theta_i \tag {say}$$ thus by power mean inequality $$\cos \theta_i\ge \sqrt{\frac{S_i^2}{9}}=\frac{S_i}{3}$$ Hence $$\dfrac{\cos ... 4 This is a remark on Dr. Mathva's answer, which may seem very mysterious at first but becomes more transparent when you consider a simpler problem: If a, b \geq 0 with a^2 + b^2 = 1, prove that$$ ab+1 \geq a+b \,.(It is the problem from the question if you impose c = d = 1.) Here it is clear that you can factorize it as (a-1)(b-1) \geq 0, so you ... 4 The method suggested by @Albus Dumbledore can be tweaked slightly to prove that the upper bound of the given expression is 9. Note that, since \theta_i \in \left[0,\dfrac{\pi}{2}\right], \cos(\theta_i) \geq 0 and \sin(\theta_i) \geq 0. We will show that \cos(\theta_i) \leq S_i: \begin{align} \cos(\theta_i) \leq S_i & \iff \cos^2(\theta_i) \leq ... 4 First, you need to prove that a_n converges to 0. With what you wrote, you can notice that \forall n, a_{n+1}\leq a_n  and a_n>0 . Hence, a_n is monotone and bounded, so it converges. Since you have f(a_n)=a_{n+1}  with f the bijection of \frac{2x}{1+x^2} , it must converge to a fixed point of f, and you can notice that 0 is one. (edit ... 4 Using the double-angle formula for the hyperbolic tangent one can obtain an explicit formula for a_n and get the desired inequality: We know that a_n \in (0, 1), so that we can substitute a_n = \tanh(x_n) with x_n > 0. Then \tanh(x_n) = \frac{2 \tanh(x_{n+1})}{1+\tanh^2(x_{n+1})} = \tanh(2 x_{n+1}) $$implies x_n = 2x_{n+1}. It follows that ... 3 Showing that a+b > - \frac73:$$(a-2)^2+(b-2)^2+(a-b)^2\geq(a-2)^2+(b-2)^2 \geq\frac{(a+b)^2}{2}-4(a+b)+8$$Now using the previous result a + b < -2 we get:$$\frac{(a+b)^2}{2}-4(a+b)+8> 18$$From OP's equation in the question, this gives$$a+b=-\frac{6}{(a-2)^2+(b-2)^2+(a-b)^2}-2>-\frac{6}{18} -2 = -\frac{7}{3}. \qquad \Box$$3$$ |z| = \sqrt{Re(z)^2 + Im(z)^2}  \leq \sqrt{Re(z)^2} + \sqrt{Im(z)^2} = |Re(z)| + |Im(z)|$$The inequality follows since in general for any non-negative x and y we have \sqrt{x+y} \leq \sqrt{x} + \sqrt{y} (consider the square of both sides) 3 Let A(x,y,z)=(1+x^2)(1+y^2)(1+z^2), defined on \{(x,y,z)\in\mathbb{R}^3{\,\mid\,}x+y+z=3\}. Since \max(|x|,|y|,|z|)\ge 3 would yield A > 8, it follows A has a global minimum value, a say. Since (x,y,z)=1 yields A=8, it follows that a\le 8. Suppose A(x,y,z)=a. Without loss of generality, assume x\le y\le z. From x+y+z=3, it ... 3 With Lagrange multipliers: consider f(x,y,z)=(x^2+1)(y^2+1)(z^2+1) and$$ L(x,y,z,t)=f(x,y,z)-t(x+y+z-3) Then you have \begin{align} \frac{\partial L}{\partial x}&=2x(y^2+1)(z^2+1)-t \\[1ex] \frac{\partial L}{\partial x}&=2y(x^2+1)(z^2+1)-t \\[1ex] \frac{\partial L}{\partial x}&=2z(x^2+1)(y^2+1)-t \end{align} These should vanish together, so ... 3 Since the function f(x)=\frac{x+2}{x^2+x} is decreasing for all x\ge 1 we haveg(x)=\frac{3^x+2}{3^{2x}+3^x} \implies g(x) \le g(1) =5/12 \space \forall x\in[1,3]$$thus$$0\le \int_1^2\frac{3^x+2}{3^{2x}+3^x}\le \frac{5}{12}(2-1)< \frac{1}{2}$$3$$f(x) = \frac {3^x + 2}{3^{2x} + 3^x}  f'(x) = -\dfrac{\ln\left(3\right)\left(3^{2x}+4{\cdot}3^x+2\right)}{3^x\left(3^x+1\right)^2} < 0 ~\forall ~ x \in \mathbb{R}$$Thus, f(x) is a decreasing function.$$ \implies f(1) >f(x) > f(2)\implies \dfrac{5}{12} >f(x) > \dfrac{11}{90}\implies \int_1^2 \dfrac{5}{12} dx > \int_1^2 ...

3

$$(x+y) (y+z) = 2\csc \theta$$ So what is the minimum value of $2\csc \theta$? What is the corresponding value of $\theta$? What kind of relationship does this indicate among $x, y, z$?

3

partial proof when $a,b,c\ge -1$ WLOG $c=\min(a,b,c)\implies 3c-13<0$ $$|1-abc|+|1-ab-bc-ca|\ge |2-abc-ab-bc-ca|\ge 2-abc-ab-bc-ca$$ It remains to prove $$2-abc-ab-bc-ca\ge 44/27$$ $$\iff 2-ab(1+c)-c(a+b)-44/27\ge 0$$ $$2-\frac{{(a+b)}^2}{4}(1+c)-c(1-c)-44/27\ge 0$$ $$\iff 2-\frac{{(1-c)}^2}{4}(1+c)-c(1-c)-44/27\ge 0$$ $$\iff \frac{-1}{108}(3c-13){(3c-1)}... 3 Let p = a + b + c = 1, q = ab + bc + ca and r = abc. Fact 1: q^2 \ge 3pr. (The proof is given at the end.) Fact 2: p^2 \ge 3q. (Proof: p^2 - 3q = \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2] \ge 0.) We split into four cases: q \ge 0: By Fact 2, we have q \le \frac{1}{3}. By Facts 1-2, we have r\le \frac{1}{27}. We have |1- q| + |1 - r| = 1 - ... 2 No calculus is required. Using your substitution in your linked effort: Let x=\frac{1-p}{p} then p=\frac{1}{1+x}, x>0. The inequality becomes$$\frac{1}{1+x}+\frac{1}{1+x}\cdot \frac{1-x}{x^n-1} = \frac{x-x^n}{(1+x)(1-x^n)} \le \frac1 2-\frac{1}{2n}\iff g(x) = \frac{x-x^n}{(1+x)(1-x^n)} \le \frac 12 - \frac{1}{2n}\tag 1$$From here, notice that ... 2 Hint: Let x=a+b then b=x-a so we get an quadratic equation on a:$$3a^2(x+2)-3a(x^2+2x)+x^3+11=0$$Since a is real it discriminant is non negative, so we have$$-3(x+2)(x^3-6x^2+44)\geq 0$$2$$a^3-6 a b+b^3=-11$$can be written as$$(a+b)^3-3 a b (a+b)-6 a b=-11$$setting$$a+b=s;\;ab=p$$we get$$ s^3-3ps-6p+11=0$$The equation$$z^2-sz+p=0\tag{1}$$Gives the values of a,b. In order to have real roots, discriminant must be positive$$s^2-4p\ge 0$$Solve$$\begin{cases} s^3-3ps-6p+11=0\\ s^2-4p\ge 0\\ \end{cases} $$Solving the first equation ... 2 As you stated$$\left( \left( \frac{\sqrt2}a\right)^2 + \left(\frac1b\right)^2\right)\left( a^2+b^2\right) \ge \left(\sqrt2 + 1 \right)^2$$Now, let's check if it can be attained. If b=1 and a=\sqrt[4]2,$$\left( \left( \frac{\sqrt2}{\sqrt[4]2}\right)^2 + \left(\frac11\right)^2\right)\left( \sqrt[4]2^2+1^2\right) = \left(\sqrt2 + 1 \right)^2$$2 Just expand to find$$2+1+\dfrac{2b^2}{a^2}+\dfrac{a^2}{b^2}=3+\left(\dfrac{\sqrt2b}a-\dfrac ab\right)^2+2\sqrt2\ge3+2\sqrt2 the equality occurs if $\dfrac{\sqrt2b}a-\dfrac ab\iff\sqrt2b^2=a^2$ with $a,b$ non zero finite real

2

AM-GM $3+2(b^2/a^2)+(a^2/b^2) \ge$ $3+2\sqrt{2(b^2/a^2)(a^2/b^2)}=$ $3+2\sqrt{2};$ Equality: $2(b^2/a^2)=(a^2/b^2);$ $2b^4=a^4,$ or $a^2=\sqrt{2}b^2.$

2

Hint: Expand the product of $y$ and $a+b$: Do you know AM-HM?

2

Since $b=2-a$, we must minimize $f(a)=\dfrac 1a+\dfrac4{2-a}$, which is a function of one variable. At a minimum, the derivative $f'(a)=-\dfrac1{a^2}+\dfrac4{(2-a)^2}$ is zero. Thus, $\dfrac 4{(2-a)^2}=\dfrac1{a^2}$ or $3a^2+4a-4=(3a-2)(a+2)=0$. Can you solve for $a$ and find $f(a)$? It is a minimum, because the second derivative \$f''(a)=\dfrac2{a^3}+\...

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