# Hung Viet's Inequality IV

### Solution 1

\displaystyle \begin{align} \frac{1}{x+5y}+\frac{1}{y+2z+3x} &\ge \frac{(1+1)^2}{4x+6y+2z}\\ &=\frac{2}{z+2x+3y}. \end{align}

It follows that

\displaystyle \begin{align} &\sum_{cycl}\frac{1}{a+5b}+\sum_{cycl}\frac{1}{b+2c+3a}\ge\sum_{cycl}\frac{2}{c+2a+3b}&\,\Leftrightarrow\\ &\sum_{cycl}\frac{1}{a+5b}\ge\sum_{cycl}\frac{1}{a+2b+3c}. \end{align}

### Solution 2

Let's introduce $a'=a+5b,\,$ $b'=b+5c,\,$ $c'=c+5a.\,$ Then

\begin{align} a'-5b'+25c'&=(a+5b)-(5b+25c)+(25c+5a)\\ &=126a. \end{align}

Thus $\displaystyle a=\frac{1}{126}(a'-5b'+25c').$ Express $b$ and $c$ similarly. To continue,

\displaystyle \begin{align} a+2b+3c&=\small{\frac{1}{126}[(a'-5b'+25c')+2(b'-5c'+25a')+3(c'-5a'+25b')]}\\ &=\frac{1}{126}[36a'+72b'+18c']\\ &=\frac{1}{7}[2a'+4b'+c']. \end{align}

It follows by Jensen's inequality that

\displaystyle \begin{align} \sum_{cycl}\frac{7}{2a'+4b'+c'}&\le\sum_{cycl}\frac{2}{7a'}+\sum_{cycl}\frac{4}{7b'}+\sum_{cycl}\frac{1}{7c'}\\ &=\sum_{cycl}\frac{2}{7a'}+\sum_{cycl}\frac{4}{7a'}+\sum_{cycl}\frac{1}{7a'}\\ &=\sum_{cycl}\frac{1}{a'}\\ &=\sum_{cycl}\frac{1}{a+5b}. \end{align}

### Solution 3

Let $b =a+\epsilon ,c=a+\xi +\epsilon ,\epsilon ,\xi \gt 0.$ Then

\displaystyle \begin{align} \text{lhs}&=\frac{1}{6 a+5 \epsilon } +\frac{1}{6 a+\xi +\epsilon } +\frac{1}{6 a+5 \xi +6 \epsilon }\\ \text{rhs}&=\frac{1}{6 a+3 \xi +5 \epsilon }+\frac{1}{6 a+2 \xi +3 \epsilon }+\frac{1}{6 a+\xi +4 \epsilon }\\ \frac{1}{6 a+5 \epsilon }&= \frac{1}{6 a+3 \xi +5 \epsilon }+\frac{3 \xi }{(6 a+5 \epsilon ) (6 a+3 \xi +5 \epsilon )}\\ \frac{1}{6 a+\xi +\epsilon }&=\frac{1}{6 a+2 \xi +3 \epsilon }+\frac{\xi +2 \epsilon }{(6 a+\xi +\epsilon ) (6 a+2 \xi +3 \epsilon )}\\ \frac{1}{6 a+5 \xi +6 \epsilon }&= \frac{1}{6 a+\xi +4 \epsilon }-\frac{2 (2 \xi +\epsilon )}{(6 a+\xi +4 \epsilon ) (6 a+5 \xi +6 \epsilon )} \end{align}/p>

We have, collecting the $\epsilon$:

$\displaystyle \frac{2 \epsilon }{(6 a+\xi +\epsilon ) (6 a+2 \xi +3 \epsilon )}\geq \frac{2 \epsilon }{(6 a+\xi +4 \epsilon ) (6 a+5 \xi +6 \epsilon )}$

And we have left to prove, collecting the $\xi.$ Let $A=6 a$

$\displaystyle \frac{3}{(A+5 \epsilon ) (A+3 \xi +5 \epsilon )}+\frac{1}{(A+\xi +\epsilon ) (A+2 \xi +3 \epsilon )}\geq \frac{4}{(A+\xi +4 \epsilon ) (A+5 \xi +6 \epsilon )}$

We can use brute force, taking the lowest common multiple, $\displaystyle Y=\frac{Y_1}{Y_2}$:

\displaystyle \begin{align} Y_1&=A^3 (2 \xi +\epsilon )+A^2 \left(9 \xi ^2+13 \epsilon ^2+23 \xi \epsilon \right)+A \left(12 \xi ^3+58 \epsilon ^3+98 \xi \epsilon ^2+57 \xi ^2 \epsilon \right)\\ &\qquad\qquad+5 \xi ^4+86 \epsilon ^4+154 \xi \epsilon ^3+99 \xi ^2 \epsilon ^2+31 \xi ^3 \epsilon \\ Y_2&=\frac{1}{6}(A+5 \epsilon ) (A+\xi +\epsilon ) (A+\xi +4 \epsilon ) (A+2 \xi +3 \epsilon ) (A+3 \xi +5 \epsilon ) (A+5 \xi +6 \epsilon ) \end{align}

Obviously $Y_1\geq0$.

### Solution 4

Denote $x=a+2b+3c,y=b+2c+3a,z=c+2a+3b$. Then the inequality becomes

$\displaystyle \frac{1}{2x-y}+\frac{1}{2y-z}+\frac{1}{2z-x}\geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$

where denominators are positive. Assume $x\geq y\geq z.$

Let

$f\displaystyle (x)=\frac{1}{2x-y}+\frac{1}{2y-z}+\frac{1}{2z-x}-\frac{1}{x}-\frac{1}{y}-\frac{1}{z}.$

Then

$\displaystyle f'(x)=\frac{-2}{(2x-y)^2}+\frac{1}{(2z-x)^2}+\frac{1}{x^2}.$

We have $0 < x\leq 2x-y,0 < 2z-x\leq 2x-y$, hence

$\displaystyle \frac{1}{x^2}\geq \frac{1}{(2x-y)^2},\frac{1}{(2z-x)^2}\geq \frac{1}{(2x-y)^2}$,

so $\displaystyle f'(x)\geq 0$. Then

$\displaystyle f(x)\geq g(y):=f(y)=\frac{1}{2y-z}+\frac{1}{2z-y}-\frac{1}{y}-\frac{1}{z}\\ \displaystyle g'(y)=\frac{-2}{(2y-z)^2}+\frac{1}{(2z-y)^2}+\frac{1}{y^2}$.

Since $0 \lt y\leq 2y-z,$ $0 \lt 2z-y\leq 2y-z$, $g'(y)\geq 0$, so $f(x)\geq g(y)\geq g(z)=0$.

Note: A similar proof shows that

$\displaystyle \frac{1}{Mx-Ny}+\frac{1}{My-Nz}+\frac{1}{Mz-Nx}\geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$

where denominators are positive and $M-N=1.$

### Acknowledgment

I lifted this inequality from the Solving the Inequality ***** facebook group. The author - Nguyen Viet Hung - remarked that the inequality is classic. Solution 1 is by Phan Ngoc Chau; Solution 2 is by @Plutoo on twitter; Solution 3 is by N. N. Taleb; Solution 4 is by Bogdan Lataianu.