how can a prove that at least one of those is less than or equal to 1/4. $$\forall a,b,c\in \mathbb R^+, \ a(1-b)\leq 1/4 \lor b(1-c) \leq 1/4 \lor c(1-a) \leq 1/4.$$ help please!

In Spivak Calculus you are asked to prove that in Schwarz inequality, equality holds only when $y_1 = y_2 = 0$ or when there is a number $\lambda$ such that $x_1 = \lambda y_1$ and $x_2 = \lambda y_2$. I can go from $x_1y_1 + x_2y_2 = \sqrt{x_1^2 + x_2^2} + \sqrt{y_1^2 + y_2^2}$ to […]

I’ve got a few questions about the problem. Prob :Suppose $(s_n)$ converges and that $s_n \geq a$ for all but finitely many terms, show $\lim s_n \geq a$ The solution here breaks this problem up into two parts. Q1. I don’t understand why is it necessary to consider the finitely many terms that $s_k < […]

Show that $$\frac xy + \frac yz + \frac zx \ge 1 + \frac {z + x}{x + y} + \frac {x + y}{z + x}$$ for $x, y, z \gt 0$. I observed that this is a homogeneous inequality so normalization might work. I tried to set $x = 1$ or $xyz = 1$ […]

How do you prove the following without induction: 1)$\prod\limits_{k=1}^n\left(\frac{2k-1}{2k}\right)^{\frac{1}{n}}>\frac{1}{2}$ 2)$\prod\limits_{k=1}^n \frac{2k-1}{2k}<\frac{1}{\sqrt{2n+1}}$ 3)$\prod\limits_{k=1}^n2k-1<n^n$ I think AM-GM-HM inequality is the way, but am unable to proceed. Any ideas. Thanks beforehand.

Given $\sin^2\alpha+\sin^2\beta+\sin^2\gamma=2 $. I have to prove that $ \left| \begin{matrix} \cos\alpha & \cos\beta & \sin\gamma\\\sin\alpha & \cos\beta & \cos\gamma\\\cos\alpha & \sin\beta & \cos\gamma \end{matrix} \right| \leq 2\sqrt2 \sin\alpha \sin\beta \sin\gamma $. I decided to directly expand the determinant, the left becomes $ |2\cos \alpha\cos\beta\cos\gamma-\sin(\alpha+\beta+\gamma)| $. This is quite different from what I encountered before […]

Let $n\in\mathbb{N}$. For $0\le l\le n$ consider \begin{equation} b_l:=4^{-l} \sum_{j=0}^l \frac{\binom{2 l}{2 j} \binom{n}{j}^2}{\binom{2 n}{2 j}}\text{.} \end{equation} Do you know a technique how to prove that \begin{equation} b_l\ge b_n\text{,$\quad 0\le l\le n-1$?} \end{equation} Going through a long list of binomial identities I did not find epiphany. Addition: Plot of $b_l$ for $n=20$.

I have the following conjecture: \begin{equation} \text{Re}\left[(1+\text{i}y)\arctan\left(\frac{t}{1+\text{i}y}\right)\right] \ge \arctan(t), \qquad \forall y,t\ge0. \end{equation} Which seems to be true numerically. Can anyone offer some advice on how to approach proving (or disproving) this? It originates from a question involving the (complex) Hilbert transform of a symmetric non-increasing probability distribution: \begin{equation} h(y) = (1+\text{i}y)\int_{-\infty}^\infty \frac{1}{1 + \text{i}(y-t)}\text{d}G(t) […]

Is there a name for the “famous” inequality $1+x \leq e^x$? It has many variants depending on how you arrange the terms: $$1 + x \leq e^x$$ $$e^{-x} -x – 1 \geq 0 $$ $$\ln(1+x) \leq x$$ Et cetera. Perhaps the simplest mnemonic device is, “$e^x$ lies above its tangent line at the origin.” This […]

Let $x, \ y, \ p$ be any real numbers with $x>0$, $y>0$, and $p>1$. The question is about (most probably) an elementary inequality: Is it always true that $x^p+y^p\leq (x+y)^p$ ? Note that if $p$ is any positive integer, then the above inequality is obviously correct. What about if the number $p \ (\text{with} […]

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