Let $a,b,c$ be positive real numbers. Prove the inequality $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2}.$$
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.
Wednesday, 9 September 2015
Tuesday, 8 September 2015
Even extension
Find the even extension of $$f(x)=\begin{align}\begin{cases}x^2-x^3,&\quad 0\leq x<3\\4-x,&\quad x\geq 3.\end{cases}\end{align}$$
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 $$g(x)=\begin{align}\begin{cases}x^2+x^3,&\quad -3<x\leq 0\\4-|x|,&\quad x>3 \vee x<-3\\x^2-x^3,&\quad 0\leq x<3.\end{cases}\end{align}$$
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 $$g(x)=\begin{align}\begin{cases}x^2+x^3,&\quad -3<x\leq 0\\4-|x|,&\quad x>3 \vee x<-3\\x^2-x^3,&\quad 0\leq x<3.\end{cases}\end{align}$$
Monday, 7 September 2015
A problem on minimisation
Find $$\min \bigg(\sqrt[4]{\frac{a}{b+c}}+\sqrt[4]{\frac{b}{a+c}}+\sqrt[4]{\frac{c}{b+a}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{a+c}{b}}+\sqrt{\frac{a+b}{c}}\bigg)\quad a,b,c\in\Bbb R^+.$$
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 $\frac{1}{3}\bigg(f(a)+f(b)+f(c)\bigg)\geq f\bigg(\frac{1}{3}(a+b+c)\bigg)$: $$\begin{align}\sqrt[4]{\frac{a}{b+c}}+\sqrt{\frac{b+c}{a}}+\sqrt[4]{\frac{b}{a+c}}+\sqrt{\frac{a+c}{b}}+\sqrt[4]{\frac{c}{b+a}}+\sqrt{\frac{a+b}{c}}&\geq 3f(\frac{1}{3})\\&= 3\bigg((\frac{1}{2})^{\frac{1}{4}}+2^{\frac{1}{2}}\bigg).\end{align}$$
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
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
Sunday, 6 September 2015
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 $x+y+z=1$.
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 $x=\dfrac{1}{11},y=\dfrac{3}{11},z=\dfrac{7}{11}$.
Today's problem:
Find the maximum of $xy^3z^7$ with non-negative real numbers in the plane $x+y+z=1$.
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 $x=\dfrac{1}{11},y=\dfrac{3}{11},z=\dfrac{7}{11}$.
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