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The dot product and Euclidean norm are $\mathbf{x}\cdot \mathbf{y}={\sum }_i{x}_i{y}_i$ and $|\mathbf{x}|=\sqrt{\mathbf{x}\cdot \mathbf{x}}$. For every real $t$,Its quadratic discriminant is nonpositive, proving the Cauchy-Schwarz inequality $|\mathbf{x}\cdot \mathbf{y}|\le |\mathbf{x}||\mathbf{y}|$. In the inequality asked for, equality holds exactly when $\mathbf{x}$ is a positive scalar multiple of $\mathbf{y}$, and

Using the Levi-Civita symbol and its contraction identity,

Put $T=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$, the scalar triple product. Cyclic symmetry gives $\mathbf{m}\cdot \mathbf{a}=\mathbf{m}\cdot \mathbf{b}=\mathbf{m}\cdot \mathbf{c}=T$, so each requested angle $\varphi$ satisfies $\cos \varphi =T/|\mathbf{m}|$. If every pairwise angle is $\theta$ and $t=\cos \theta$, the preceding identity givesand therefore $|\mathbf{m}|=\sqrt{3}(1-\cos \theta )$. This is $0$ when $\theta =0$; for a right-handed orthonormal basis and $\theta =\pi /2$, $\mathbf{m}=\mathbf{a}+\mathbf{b}+\mathbf{c}$ has norm $\sqrt{3}$.

The three planes have equations $\mathbf{a}\cdot \mathbf{x}=\mathbf{a}\cdot \mathbf{p}$, $\mathbf{b}\cdot \mathbf{x}=\mathbf{b}\cdot \mathbf{q}$, and $\mathbf{c}\cdot \mathbf{x}=\mathbf{c}\cdot \mathbf{r}$. They meet at one point exactly when $\Delta =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$, meaning their normals are linearly independent vectors. The point is

The radius of convergence of the power series ${\sum }_{n\ge 0}{a}_n{z}^n$ isIf $|z|<R$, choose $w$ with $|z|<|w|<R$. Since $a_n{w}^n\to 0$, those terms are bounded, say $|a_n{w}^n|\le M$. Thereforeand comparison with a geometric series proves absolute convergence. If $|z|>R$, convergence would imply absolute convergence at every smaller modulus by the same argument, contradicting the definition of $R$.

If $|a_n|\le |b_n|$, convergence of $\sum b_n{w}^n$ implies absolute convergence of $\sum a_n{z}^n$ whenever $|z|<|w|$. Letting $|w|$ approach the radius of the $b$-series shows that $R_a\ge R_b$.

Let $(u_n)$ be increasing and bounded above, and let $L=\sup \{u_n:n\in \mathbb{N}\}$. For every $\epsilon >0$, the definition of the supremum supplies $N$ with $L-\epsilon <u_N\le L$. Monotonicity gives $L-\epsilon <u_n\le L$ for every $n\ge N$, proving $u_n\to L$. This is the monotone bounded sequence theorem.

For the recurrence, $f(x_n)>0$ makes $(x_n)$ increasing. If it were bounded above, it would converge to some $L$. Since $x_n\le L$ and $f$ is decreasing, $f(x_n)\ge f(L)>0$, whencea contradiction. Therefore $x_n\to \infty$.

The comparison test for series states that if $0\le a_n\le b_n$ eventually and $\sum b_n$ converges, then $\sum a_n$ converges. The partial sums of $\sum a_n$ are increasing and bounded above by a constant plus the bounded partial sums of $\sum b_n$, so the monotone bounded sequence theorem applies.

If $\sum x_n$ converges, then $x_n^2\le x_n$ proves convergence of $\sum x_n^2$. Also $x_n\to 0$, so eventually $x_n\le 1/2$ and $x_n/(1-x_n)\le 2x_n$, proving convergence of $\sum x_n/(1-x_n)$. The converse for the square series is false: $x_n=1/(n+1)$ has convergent square series but divergent original series. The converse for the fraction series is true because $x_n\le x_n/(1-x_n)$.

If $(x_n)$ is positive and decreasing and $\sum x_n$ converges, put $m=\lfloor n/2\rfloor$. Thenso $n{x}_n\to 0$. The converse fails for $x_n=1/(n\log n)$ for $n\ge 2$: it is decreasing and $n{x}_n\to 0$, but the integral test for convergence shows that its series diverges.

Even convergence of $\sum x_n$ need not imply $(n\log n)x_n\to 0$. Choose integers $N_j=\lceil e^{2^j}\rceil$ and set $x_n=(N_j\log N_j{)}^{-1}$ for $N_{j-1}<n\le N_j$. This is decreasing, and the $j$th block contributes at most $1/\log N_j\le 2^{-j}$, so the series converges. At $n=N_j$, however, $(n\log n)x_n=1$.