$|z-1|\leq \bigg| |z|-1\bigg|+|z| |Arg\:z|$

For a non-zero $z$ and $-\pi<Arg\:z\leq \pi$, show that $$|z-1|\leq \bigg| |z|-1\bigg|+|z| |Arg\:z|.$$

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$$\begin{align}|z-1|&=\bigg|z-|z|+|z|-1\bigg|\\
&\leq \bigg|z-|z|\bigg|+\bigg||z|-1\bigg|\\
&=|z||\cos\theta+i\sin\theta-1|+\bigg||z|-1\bigg|\\
&=|z|\sqrt{(\cos\theta-1)^2+\sin^2\theta}+\bigg||z|-1\bigg|\\
&=|z|\left|2\sin\frac{\theta}{2}\right|+\bigg||z|-1\bigg|\\
&\leq |z||Arg\:z|+\bigg||z|-1\bigg|\end{align}$$ since $(\cos\theta-1)^2+\sin^2\theta=(1-2\sin^2\dfrac{\theta}{2}-1)^2+4\sin^2\dfrac{\theta}{2}\cos^2\dfrac{\theta}{2}=4\sin^2\dfrac{\theta}{2}$ and $\bigg|\sin\dfrac{\theta}{2}\bigg|<\bigg|\dfrac{\theta}{2}\bigg|$.

Geometric interpretation

Take $|z|>1$. Consider the triangle with vertices $1,z,|z|$. By the triangle inequality, we have $|z-1|\leq \bigg|z-|z|\bigg|+\bigg||z|-1\bigg|$. Recall the arc length formula $s=r\theta$. Here, $|z| |Arg\:z|$ is actually the arc length joining $z$ and $|z|$, which is larger than the chord joining the two points. Therefore, the inequality holds.

Discovery work in analysis

In this post, we will deal with statements of the form $$\forall x\;\exists y \bullet P(x,y).$$ To prove these statements, we need some preliminary 'discovery work' before embarking on the proof.

Convergence of sequences
Definition: The sequence $(a_n)$ converges to $l$ if, for any $\epsilon>0$, there exists $N\in \Bbb Z^+$ such that $$n>N \Rightarrow |a_n-l|<\epsilon.$$ We can write either $a_n\to l$ as $n\to \infty$ or $\lim\limits_{n\to \infty}a_n=l$.

Example
A sequence $(a_n)$ is defined by $a_n=\dfrac{3n^2-4n}{(n+1)(n+2)}$ for $n\in \Bbb Z^+$. Show that $\lim\limits_{n\to \infty}a_n=3$.

The definition of convergence involves two quantifiers and is of the form $$\forall \epsilon\;\exists N \bullet P(\epsilon,N)$$ where $P(\epsilon,N)$ is a propositional function and the universes for $\epsilon$ and $N$ are $\Bbb R^+$ and $\Bbb Z^+$ respectively. The structure of the proof starts with an arbitrary $\epsilon$ (in its universe) and then selects a particular $N$ (in its universe) which may depend on $\epsilon$. To complete the proof, we need to show that the propositional function $P(\epsilon,N)$ is satisfied for the arbitrary $\epsilon$ and particular $N$.

We first start by the definition. We need to find a positive integer $N$ such that for all $n>N$, $$|a_n-l|=\left|\frac{3n^2-4n}{(n+1)(n+2)}-3 \right|<\epsilon.$$ But how? We can manipulate $|a_n-l|$ until it is less than any positive $\epsilon$: $$\begin{align}\left| \dfrac{3n^2-4n}{(n+1)(n+2)}-3 \right|&=\left| \dfrac{3n^2-4n-3(n+1)(n+2)}{(n+1)(n+2)} \right|\\&=\left| \dfrac{3n^2-4n-(3n^2+9n+6)}{(n+1)(n+2)} \right|\\&=\left| \dfrac{-13n-6}{(n+1)(n+2)} \right|\\&=\dfrac{13n+6}{(n+1)(n+2)}.\end{align}$$ We want to show $\dfrac{13n+6}{(n+1)(n+2)}$ is less than or equal to some simple fraction which can still be made less than $\epsilon$. Since replacing the numerator with a larger value and replacing the denominator with a smaller value both increases the size of the expression respectively, we have $$\dfrac{13n+6}{(n+1)(n+2)}\leq \dfrac{19n}{(n+1)(n+2)}\leq \dfrac{19n}{n^2}=\dfrac{19}{n}.$$ Now to ensure $\dfrac{19}{n}<\epsilon$, we can take $n>\dfrac{19}{\epsilon}$. But we are not done yet. Recall that the definition requires $N$ to be a positive integer. To fix this, we take $N$ to be the integer part or floor of $\dfrac{19}{\epsilon}$, which is defined to be the largest integer less than or equal to $\dfrac{19}{\epsilon}$ and denoted by $\lfloor \dfrac{19}{\epsilon} \rfloor$. We can now proceed to the proof.

Proof: Let $\epsilon>0$ and $N=\lfloor \dfrac{19}{\epsilon}\rfloor \in \Bbb Z^+$. Then for all integers $n>N$, we have $$\begin{align}\left| \dfrac{3n^2-4n}{(n+1)(n+2)}-3 \right|&=\left| \dfrac{3n^2-4n-3(n+1)(n+2)}{(n+1)(n+2)} \right|\\&=\left| \dfrac{3n^2-4n-(3n^2+9n+6)}{(n+1)(n+2)} \right|\\&=\left| \dfrac{-13n-6}{(n+1)(n+2)} \right|\\&=\dfrac{13n+6}{(n+1)(n+2)}\\ &\leq \dfrac{13+6n}{(n+1)(n+2)}\\&\leq \dfrac{19n}{n^2}\\&=\dfrac{19}{n}\\&< \epsilon\quad \text{since}\;n>\dfrac{19}{\epsilon}.\end{align}$$ Therefore $\lim\limits_{n\to \infty}\dfrac{3n^2-4n}{(n+1)(n+2)}=3$.

Limits of functions
Definition: Let $f:A\subset \Bbb R \to B\subset \Bbb R$. Let $a\in A$. Then $f(x)$ tends to a limit $l$ as $x$ tends to $a$ if, for any $\epsilon>0$, there exists $\delta>0$ such that $$0<|x-a|<\delta \Rightarrow |f(x)-l|<\epsilon.$$ We can write either $f(x)\to l$ as $x\to a$ or $\lim\limits_{x\to a}f(x)=l$.

Example
Show that $\lim\limits_{x\to 2}2x^2-5x=-2$.

We need to ensure that $|f(x)-l|=|2x^2-5x+2|$ is less than any specified positive $\epsilon$ by taking $x$ such that $0<|x-2|<\delta$. So we consider $|2x^2-5x+2|$: $$|2x^2-5x+2|=|(x-2)(2x-1)|=|x-2||2x-1|.$$ We can control the size of $|x-2|$ since we can choose $\delta$ such that $|x-2|<\delta$. As for $|2x-1|$, we can rewrite it in terms of $|x-2|$ using the triangle inequality: $$\begin{align}|2x-1|&=|2(x-2)+3|\\ &\leq |2(x-2)|+|3|\\ &=2|x-2|+3.\end{align}$$ Therefore $$|2x^2-5x+2|=|x-2||2x-1|\leq |x-2|(2|x-2|+3).$$ Say $|x-2|<1$. Then $2|x-2|+3<5$. If, in addition, $|x-2|<\dfrac{\epsilon}{5}$, then $|x-2|(2|x-2|+3)$ will be less than $\epsilon$. We are now ready to start the proof.

Proof: Let $\epsilon>0$ and $\delta=\min\{1,\dfrac{\epsilon}{5}\}$. Then $$0<|x-2|<\delta \Rightarrow |x-2|<1 \wedge |x-2|<\frac{\epsilon}{5}\\ \begin{align}\Rightarrow |2x^2-5x+2|&=|x-2||2x-1|\\ &\leq |x-2|(2|x-2|+3)\\ &<5|x-2|\\ &<\epsilon \end{align}$$ Therefore $\lim\limits_{x\to 2}2x^2-5x=-2$.

Remark: The choice of $\delta=\min\{1,\dfrac{\epsilon}{5}\}$ is arbitrary. Choosing $\delta=\min\{\dfrac{1}{2},\dfrac{\epsilon}{4}\}$ will also work.

Convergence of $\sum_{n=2}^\infty \frac{1}{n\ln n(\ln(\ln n))^2}$

Prove the series $$\sum_{n=2}^\infty \frac{1}{n\ln n(\ln(\ln n))^2}$$ converges.

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Since the general term is a positive and decreasing sequence, by the integral test, the series behaves as the following integral: $$\int_3^\infty \frac{dx}{x\ln x (\ln(\ln x))^2}.$$By a change of variable $$u=\ln(\ln x),\quad du=\frac{dx}{x\ln x},$$ we have $$\int_3^\infty \frac{dx}{x\ln x (\ln(\ln x))^2}=\int_{\ln(\ln 3)}^\infty \frac{du}{u^2}=-\frac{1}{u}\bigg|_{\ln(\ln 3)}^{\infty}=\frac{1}{\ln(\ln 3)}.$$ So it is convergent. [We start with $n=3$ to avoid the pole at $x=e$ where the integral may diverge.]

We can also apply the Cauchy condensation test. $$\sum_n a_n\;\text{converges iff}\;\sum_n 2^n a_{2^n}\;\text{converges}.$$ Setting $a_n=\dfrac{1}{n\ln n(\ln(\ln n))^2}$, we have $$\sum_n 2^n a_{2^n}=\sum_{n=2}^\infty \frac{2^n}{2^n(n\ln 2)(\ln n+\ln(\ln 2))^2}\leq \frac{1}{\ln 2}\sum_{n=2}^\infty \frac{1}{n\ln^2 n}.$$ Similarly, for $\sum\limits_{n=2}^\infty\dfrac{1}{n\ln^2 n}$, we have $$\sum_{n=2}^\infty \frac{2^n}{2^n\ln^2(2^n)}=\frac{1}{\ln^2 2}\sum_{n=2}^\infty \frac{1}{n^2}.$$ So the series is convergent.

Find $\min x^2+y^2+z^2$

If $ax+by+cz=k$, prove that the minimum value of $x^2+y^2+z^2$ is $\dfrac{k^2}{a^2+b^2+c^2}$.

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Since $$(x^2+y^2+z^2)(a^2+b^2+c^2)-(ax+by+cz)^2=(ay-bx)^2+(bz-cy)^2+(cx-az)^2\geq 0,$$ we have $$x^2+y^2+z^2\geq \frac{(ax+by+cz)^2}{a^2+b^2+c^2}=\frac{k^2}{a^2+b^2+c^2}.$$
Alternatively, we can use geometric arguments. The normal of $ax+by+cz=k$ is $\vec{n_1}=(a,b,c)$. The normal of $x^2+y^2+z^2$ is $\vec{n_2}=(2x,2y,2z)$. Applying Cauchy Schwartz inequality for $\vec{n_1}$ and $\vec{n_2}$: $$\vec{n_1}\cdot\vec{n_2}\leq \lVert \vec{n_1}\rVert \lVert \vec{n_2}\rVert,$$ we have $$2(ax+by+cz)\leq 2\sqrt{a^2+b^2+c^2} \sqrt{x^2+y^2+z^2},$$ which means $$x^2+y^2+z^2\geq \dfrac{k^2}{a^2+b^2+c^2}.$$

Sketch $|1-3x|+|2-2x|+|3-x|$

Sketch the function $$f(x)=|1-3x|+|2-2x|+|3-x|,x\in \Bbb R.$$

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$$f(x)=|1-3x|+|2-2x|+|3-x|=|1-3x|+2|1-x|+|3-x|$$First notice that when $x<0$,$f(x)=(1-3x)+(2-2x)+(3-x)=6-6x$.

Looking at the first term (within the absolute sign), we know when $x\leq \dfrac{1}{3}$,$f(x)=6-6x$.
Now, look at the other two terms, when $1\leq x\leq 3,f(x)=-(1-3x)-2(1-x)+(3-x)=4x$.
When $x\geq 3, f(x)=-(1-3x)-2(1-x)-(3-x)=-6+6x$.
It remains to check the interval $[\dfrac{1}{3},1]$. There $f(x)=-(1-3x)+2(1-x)+(3-x)=4$.

We conclude that $$f(x)=\begin{align}\begin{cases}6-6x,\quad &x\leq \dfrac{1}{3}\\ 4,\quad &\dfrac{1}{3}\leq x\leq 1\\ 4x,\quad &1\leq x\leq 3\\ -6+6x,\quad &x\geq 3. \end{cases}\end{align}$$

Remark: If we set $f(x)$ to be zero, we can have the 'critical points' $\dfrac{1}{3},1,3$. This naturally forms four regions. $x\leq \dfrac{1}{3}, \dfrac{1}{3}\leq x\leq 1, 1\leq x\leq 3, x\geq 3$.

In fact, this problem is related to distance. The Manhattan distance $d_M$ between points $P(x_1,y_1)$ and $Q(x_2,y_2)$ in the plane is defined by $$d_M(P,Q)=|x_1-x_2|+|y_1-y_2|.$$ The $d_M$ distance is the sum of the horizontal and vertical distances between the points like 'three blocks west and two blocks north'.

Here's an extra problem that can be solved using the notion of distance.

If $a<b<c<d$ and $x\in \Bbb R$, find the least value of the function $$f(x)=|x-a|+|x-b|+|x-c|+|x-d|.$$

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It turns out we don't need to use derivatives for this type of problems. The value of $f(x)$ is the sum of the distances from $x$ to $a,b,c,d$. Imagine we start $x$ at the left of $a$, and move it to the right. Initially all the four terms are decreasing since $x$ is now closer to the other points. When $x$ passes $a$ and starts moving to $b$, the distance from $x$ to $a$ increases while that from $x$ to $b$ decreases at the same rate, so $|x-a|+|x-b|$ is constant. The distances from $x$ to $c$ and $d$ are decreasing, so $f(x)$ is still decreasing. We can treat the other intervals similarly.

Overall, the function is decreasing to the left of b, constant on $[b,c]$, and increasing to the right of $c$, so we know the minimum value is on $[b,c]$. When $b\leq x\leq c$, the sum of distances from $x$ to $a$ and $d$ is $d-a$, and that from $x$ to $b$ and $c$ is $c-b$. Therefore, the total distance is $c+d-a-b$.

More to read
The median minimises the sum of absolute deviations

$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2}$

Let $a,b,c$ be positive real numbers. Prove the inequality $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2}.$$

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From $(a-b)^2\geq 0$, we know $\dfrac{a}{b}+\dfrac{b}{a}\geq 2$. Equality occurs iff $a=b$. Using this inequality, we then have $$\frac{a+b}{b+c}+\frac{b+c}{a+b}+\frac{b+c}{c+a}+\frac{c+a}{b+c}+\frac{c+a}{a+b}+\frac{a+b}{c+a}\geq 6.$$ We rewrite the inequality as $$\bigg(\frac{a+b}{b+c}+\frac{c+a}{b+c}\bigg)+\bigg(\frac{a+b}{c+a}+\frac{b+c}{c+a}\bigg)+\bigg(\frac{b+c}{a+b}+\frac{c+a}{a+b}\bigg) \geq 6,$$ which means $$\frac{2a}{b+c}+1+\frac{2b}{c+a}+1+\frac{2c}{a+b}+1\geq 6,$$ or $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2}$$ as required.

Equality occurs iff $\dfrac{a+b}{b+c}=\dfrac{b+c}{a+b},\dfrac{b+c}{c+a}=\dfrac{c+a}{b+c},\dfrac{c+a}{a+b}=\dfrac{a+b}{c+a}$ from where we deduce $a=b=c$.

Takeaways:
1. Make use of known inequalities.
2. Reformulation of the problem may give us more insight: rewrite the inequality as $$\frac{2a}{b+c}+\frac{2b}{a+c}+\frac{2c}{a+b}\geq 3.$$ 3. We can start with $\dfrac{a+b}{b+c}+\dfrac{b+c}{a+b}\geq 2$. Since we want $\dfrac{2a}{b+c}$, we add $\dfrac{b+c}{c+a}+\dfrac{c+a}{b+c}$ to $\dfrac{a+b}{b+c}+\dfrac{b+c}{a+b}$. Note also the symmetry of the problem. We should have $2\cdot 3=6$ terms.
4. Don't forget the equality case.

Even extension

Find the even extension of $$f(x)=\begin{align}\begin{cases}x^2-x^3,&\quad 0\leq x<3\\4-x,&\quad x\geq 3.\end{cases}\end{align}$$

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We know $|x|$ is an even function. We thus have $$g(x)=\begin{align}\begin{cases}|x|^2-|x|^3,&\quad 0\leq |x|<3\\4-|x|,&\quad |x|\geq 3,\end{cases}\end{align}$$ or $$g(x)=\begin{align}\begin{cases}x^2+x^3,&\quad -3<x\leq 0\\4-|x|,&\quad x>3 \vee x<-3\\x^2-x^3,&\quad 0\leq x<3.\end{cases}\end{align}$$

A problem on minimisation

Find $$\min \bigg(\sqrt[4]{\frac{a}{b+c}}+\sqrt[4]{\frac{b}{a+c}}+\sqrt[4]{\frac{c}{b+a}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{a+c}{b}}+\sqrt{\frac{a+b}{c}}\bigg)\quad a,b,c\in\Bbb R^+.$$

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We can use AM-GM inequality just like the previous problem. Using AM-GM inequality for the first three terms with the equality case when $a=b=c$, $$\begin{align}\sqrt[4]{\frac{a}{b+c}}+\sqrt[4]{\frac{b}{a+c}}+\sqrt[4]{\frac{c}{b+a}}&\geq 3\sqrt[3]{\sqrt[4]{\frac{a}{b+c}\frac{b}{a+c}\frac{c}{b+a}}}\\&=3\bigg(\frac{1}{8}\bigg)^{\frac{1}{4}}.\end{align}$$ Similarly, for the last three terms: $$\begin{align}\sqrt{\frac{b+c}{a}}+\sqrt{\frac{a+c}{b}}+\sqrt{\frac{a+b}{c}}&\geq 3\sqrt[3]{\sqrt{\frac{b+c}{a}\frac{a+c}{b}\frac{a+b}{c}}}\\&=3(8)^{\frac{1}{6}}.\end{align}$$ Adding the two inequalities gives $$3\bigg((\frac{1}{2})^{\frac{1}{4}}+2^{\frac{1}{2}}\bigg) \leq \bigg(\sqrt[4]{\frac{a}{b+c}}+\sqrt[4]{\frac{b}{a+c}}+\sqrt[4]{\frac{c}{b+a}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{a+c}{b}}+\sqrt{\frac{a+b}{c}}\bigg).$$
Alternative solution
Let $M$ be the expression. Since $M$ is a homogeneous expression (same degree for each term), we can set $a+b+c=1$. Then we have $$M=\sum_{\text{cyc}}\bigg( \sqrt[4]{\frac{a}{1-a}}+\sqrt{\frac{1-a}{a}}\bigg).$$ Now $$f(x)=\sqrt[4]{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}$$ is a convex function. We can use Jensen's inequality $\frac{1}{3}\bigg(f(a)+f(b)+f(c)\bigg)\geq f\bigg(\frac{1}{3}(a+b+c)\bigg)$: $$\begin{align}\sqrt[4]{\frac{a}{b+c}}+\sqrt{\frac{b+c}{a}}+\sqrt[4]{\frac{b}{a+c}}+\sqrt{\frac{a+c}{b}}+\sqrt[4]{\frac{c}{b+a}}+\sqrt{\frac{a+b}{c}}&\geq 3f(\frac{1}{3})\\&= 3\bigg((\frac{1}{2})^{\frac{1}{4}}+2^{\frac{1}{2}}\bigg).\end{align}$$

Cyclic sum

A cyclic sum is a summation that cycles through all the values of a function and takes their sum.

Rigorous definition
Consider a function $f(a_1,a_2,a_3,\cdots,a_n)$. The cyclic sum $\sum\limits_{\text{cyc}}f(a_1,a_2,a_3,\cdots,a_n)$ is equal to $$f(a_1,a_2,a_3,\cdots,a_n)+f(a_2,a_3,a_4,\cdots,a_n,a_1)+\cdots+f(a_n,a_1,a_2,\cdots,a_{n-2},a_{n-1}).$$ The notation $\sum\limits_{\text{cyc}}$ implies that all variables are cycled through. Another notation is $\sum\limits_{a,b,c}$, which implies that the cyclic sum only cycle through those variables underneath the sigma. [Note: Do not confuse this notation with the symmetric sum.]

Examples
Consider the permutation $p=(a\;b\;c)$. The cyclic sum $\sum\limits_p a$ is the sum that cycles through the permutation: $$\sum_p a=a+b+c.$$
They often come up in inequalities: $$\sum_{a,b,c}\frac{a^3}{3}=\frac{a^3}{3}+\frac{b^3}{3}+\frac{c^3}{3}\geq \sqrt[3]{\frac{(abc)^3}{3^3}}=\frac{abc}{3}.$$
They are extremely helpful in inequalities involving many letters. Instead of writing all the terms of the sum explicitly, we can employ this notation. Check out this answer.

Cyclic numbers
$142857$, the six repeating digits of $\frac{1}{7}$, $0.\overline{142857}$, is the best-known cyclic number in base $10$.
$$1\times 142,857=142,857\\ 2\times 142,857=285,714\\ 3\times 142,857=428,571\\ 4\times 142,857=571,428\\ 5\times 142,857=714,285\\ 6\times 142,857=857,142\\ 7\times 142,857=999,999$$
When multiplied by $2, 3, 4, 5$, or $6$, the answer will be a cyclic permutation of itself, and will correspond to the repeating digits of $\dfrac{2}{7},\dfrac{3}{7},\dfrac{4}{7},\dfrac{5}{7},\dfrac{6}{7}$, respectively.
$$1\div 7=0.\overline{142,857}\\ 2\div 7=0.\overline{285,714}\\ 3\div 7=0.\overline{428,571}\\ 4\div 7=0.\overline{571,428}\\ 5\div 7=0.\overline{714,285}\\ 6\div 7=0.\overline{857,142}\\ 7\div 7=0.\overline{999,999}=1\\ 8\div 7=1.\overline{142,857}\\ 9\div 7=1.\overline{285,714}$$
One last interesting thing about this fraction: $$\begin{align}\frac{1}{7}&=0.142857142857...\\&=0.14+0.0028+0.000056+0.00000112+0.0000000224+0.000000000448+\cdots\\&=\frac{14}{100}+\frac{28}{100^2}+\frac{56}{100^3}+\frac{112}{100^4}+\frac{224}{100^5}+\cdots+\frac{7\cdot 2^n}{100^n}+\cdots\\&=\frac{7}{50}+\frac{7}{50^2}+\frac{7}{50^3}+\frac{7}{50^4}+\frac{7}{50^5}+\cdots+\frac{7}{50^n}+\cdots\\&=\sum_{k=1}^\infty \frac{7}{50^k}.\end{align}$$Each term is double the prior term shifted two places to the right.

Reference
Definition of cyclic sum
Examples

More to explore
The Alluring Lore of Cyclic Numbers by Michael W. Ecker

Find $\max xy^3z^7$ where $x+y+z=1$

As a brief introduction, 'problem of the day' posts will be about mathematical thinking (problem-solving and proofs). The questions and solutions are extracted from online sources. I will add alternative solutions and mention something interesting about the problems if I can.

Today's problem:

Find the maximum of $xy^3z^7$ with non-negative real numbers in the plane $x+y+z=1$.

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We make use of AM-GM inequality: $$1=x+y+z=x+\frac{1}{3}y+\frac{1}{3}y+\frac{1}{3}y+\frac{1}{7}z+\cdots+\frac{1}{7}z\geq 11 \sqrt[11]{x(\frac{1}{3}y)^3(\frac{1}{7}z)^7}.$$ To obtain the maximum, we want to find $x,y,z$ for the equality case. Namely, $x=\dfrac{1}{3}y=\dfrac{1}{7}z=\dfrac{1}{11}$. We thus have the maximum $xy^3z^7=\dfrac{3^3 7^7}{11^{11}}$.

Alternatively, we can use lagrange multiplier since this is an optimisation problem with a constraint. Let $F(x,y,z,\lambda)=xy^3z^7-\lambda(x+y+z-1)$. Then, we have a system of equations $$\begin{cases}F_x=y^3z^7-\lambda=0\\F_y=3xy^2z^7-\lambda=0\\F_z=7xy^3z^6-\lambda=0\\ x+y+z=1.\end{cases}$$ From the first three equations, we know $y=3x,z=7x$. Therefore, substituting them into the last equation, we have $x+3x+7x=1$ and thus $x=\dfrac{1}{11},y=\dfrac{3}{11},z=\dfrac{7}{11}$.