Let $(x,y,z)\in \mathbb{R}$, prove that $(}\frac{2x-y}{x-y}{{\textstyle )}}^{2}+{\textstyle (}\frac{2y-z}{y-z}{{\textstyle )}}^{2}+{\textstyle (}\frac{2z-x}{z-x}{{\textstyle )}}^{2}\u2a7e5$

Kelton Bailey
2022-09-30
Answered

Let $(x,y,z)\in \mathbb{R}$, prove that $(}\frac{2x-y}{x-y}{{\textstyle )}}^{2}+{\textstyle (}\frac{2y-z}{y-z}{{\textstyle )}}^{2}+{\textstyle (}\frac{2z-x}{z-x}{{\textstyle )}}^{2}\u2a7e5$

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smh3402en

Answered 2022-10-01
Author has **11** answers

Start with

$$\begin{array}{r}(2{y}^{2}z+2{z}^{2}x+2{x}^{2}y-y{z}^{2}-z{x}^{2}-x{y}^{2}-3xyz{)}^{2}\ge 0\end{array}$$

This can be expanded out to give

$$\begin{array}{r}4\sum _{cyc}{y}^{2}{z}^{4}-4\sum _{cyc}{y}^{3}{z}^{3}+\sum _{cyc}{y}^{4}{z}^{2}-4\sum _{cyc}xy{z}^{4}+\\ 14\sum _{cyc}x{y}^{2}{z}^{3}-10\sum _{cyc}x{y}^{3}{z}^{2}-3{x}^{2}{y}^{2}{z}^{2}\ge 0\end{array}$$

This can be rearranged to

$$\begin{array}{r}(2x-y{)}^{2}(y-z{)}^{2}(z-x{)}^{2}+(x-y{)}^{2}(2y-z{)}^{2}(z-x{)}^{2}+(x-y{)}^{2}(y-z{)}^{2}(2z-x{)}^{2}\ge 5(x-y{)}^{2}(y-z{)}^{2}(z-x{)}^{2}\end{array}$$

Now divide by

$(x-y{)}^{2}(y-z{)}^{2}(z-x{)}^{2}$ and we have

$$\begin{array}{r}\frac{(2x-y{)}^{2}}{(x-y{)}^{2}}+\frac{(2y-z{)}^{2}}{(y-z{)}^{2}}+\frac{(2z-x{)}^{2}}{(z-x{)}^{2}}\ge 5.\end{array}$$

$$\begin{array}{r}(2{y}^{2}z+2{z}^{2}x+2{x}^{2}y-y{z}^{2}-z{x}^{2}-x{y}^{2}-3xyz{)}^{2}\ge 0\end{array}$$

This can be expanded out to give

$$\begin{array}{r}4\sum _{cyc}{y}^{2}{z}^{4}-4\sum _{cyc}{y}^{3}{z}^{3}+\sum _{cyc}{y}^{4}{z}^{2}-4\sum _{cyc}xy{z}^{4}+\\ 14\sum _{cyc}x{y}^{2}{z}^{3}-10\sum _{cyc}x{y}^{3}{z}^{2}-3{x}^{2}{y}^{2}{z}^{2}\ge 0\end{array}$$

This can be rearranged to

$$\begin{array}{r}(2x-y{)}^{2}(y-z{)}^{2}(z-x{)}^{2}+(x-y{)}^{2}(2y-z{)}^{2}(z-x{)}^{2}+(x-y{)}^{2}(y-z{)}^{2}(2z-x{)}^{2}\ge 5(x-y{)}^{2}(y-z{)}^{2}(z-x{)}^{2}\end{array}$$

Now divide by

$(x-y{)}^{2}(y-z{)}^{2}(z-x{)}^{2}$ and we have

$$\begin{array}{r}\frac{(2x-y{)}^{2}}{(x-y{)}^{2}}+\frac{(2y-z{)}^{2}}{(y-z{)}^{2}}+\frac{(2z-x{)}^{2}}{(z-x{)}^{2}}\ge 5.\end{array}$$

flatantsmu

Answered 2022-10-02
Author has **1** answers

This is a special case of the following inequality:

$$\sum _{a,b,c}{\textstyle (}{\displaystyle \frac{ma-nb}{a-b}}{{\textstyle )}}^{2}\ge {m}^{2}+{n}^{2},$$

where $m,n\in \mathbb{R}$ and $a,b,c$ are distinct reals.

The proof is simply completing the square: start by writing

$$\frac{ma-nb}{a-b}}=m+(m-n){\displaystyle \frac{b}{a-b}$$

and

$$\frac{mc-na}{c-a}}=n+(m-n){\displaystyle \frac{c}{c-a}$$

and leave the middle term untouched first.

Let me know if still need further assistance.

EDIT: Completing the hint:

$$\sum _{a,b,c}{\textstyle (}{\displaystyle \frac{ma-nb}{a-b}}{{\textstyle )}}^{2}-{m}^{2}-{n}^{2}={\textstyle (}{\displaystyle \frac{mb-nc}{b-c}}{{\textstyle )}}^{2}+(m-n{)}^{2}{\textstyle (}{\displaystyle \frac{{b}^{2}}{(a-b{)}^{2}}}+{\displaystyle \frac{{c}^{2}}{(c-a{)}^{2}}}{\textstyle )}+2(m-n){\textstyle (}{\displaystyle \frac{mb}{a-b}}+{\displaystyle \frac{nc}{c-a}}{\textstyle )}=$$

$$={\textstyle (}{\displaystyle \frac{mb-nc}{b-c}}{{\textstyle )}}^{2}+(m-n{)}^{2}{\textstyle (}{\displaystyle \frac{{b}^{2}}{(a-b{)}^{2}}}+{\displaystyle \frac{{c}^{2}}{(c-a{)}^{2}}}{\textstyle )}+2(m-n){\displaystyle \frac{bc(m-n)-a(mb-nc)}{(a-b)(c-a)}}=$$

$$={\textstyle (}{\displaystyle \frac{mb-nc}{b-c}}{{\textstyle )}}^{2}+(m-n{)}^{2}{\textstyle (}{\displaystyle \frac{{b}^{2}}{(a-b{)}^{2}}}+{\displaystyle \frac{{c}^{2}}{(c-a{)}^{2}}}+{\displaystyle \frac{2bc}{(a-b)(c-a)}}{\textstyle )}-{\displaystyle \frac{2a(mb-nc)(m-n)}{(a-b)(c-a)}}=$$

$$={\textstyle (}{\displaystyle \frac{mb-nc}{b-c}}{{\textstyle )}}^{2}+(m-n{)}^{2}{\textstyle (}{\displaystyle \frac{b}{a-b}}+{\displaystyle \frac{c}{c-a}}{{\textstyle )}}^{2}-{\displaystyle \frac{2a(m-n)(mb-nc)}{(a-b)(c-a)}}=$$

$$(}{\displaystyle \frac{mb-nc}{b-c}}{{\textstyle )}}^{2}+(m-n{)}^{2}{\textstyle (}{\displaystyle \frac{a(c-b)}{(a-b)(c-a)}}{{\textstyle )}}^{2}-{\displaystyle \frac{2a(m-n)(mb-nc)}{(a-b)(c-a)}}=$$

$$={\textstyle (}{\displaystyle \frac{mb-nc}{b-c}}-{\displaystyle \frac{a(m-n)(c-b)}{(a-b)(c-a)}}{{\textstyle )}}^{2}\ge 0.$$

$$\sum _{a,b,c}{\textstyle (}{\displaystyle \frac{ma-nb}{a-b}}{{\textstyle )}}^{2}\ge {m}^{2}+{n}^{2},$$

where $m,n\in \mathbb{R}$ and $a,b,c$ are distinct reals.

The proof is simply completing the square: start by writing

$$\frac{ma-nb}{a-b}}=m+(m-n){\displaystyle \frac{b}{a-b}$$

and

$$\frac{mc-na}{c-a}}=n+(m-n){\displaystyle \frac{c}{c-a}$$

and leave the middle term untouched first.

Let me know if still need further assistance.

EDIT: Completing the hint:

$$\sum _{a,b,c}{\textstyle (}{\displaystyle \frac{ma-nb}{a-b}}{{\textstyle )}}^{2}-{m}^{2}-{n}^{2}={\textstyle (}{\displaystyle \frac{mb-nc}{b-c}}{{\textstyle )}}^{2}+(m-n{)}^{2}{\textstyle (}{\displaystyle \frac{{b}^{2}}{(a-b{)}^{2}}}+{\displaystyle \frac{{c}^{2}}{(c-a{)}^{2}}}{\textstyle )}+2(m-n){\textstyle (}{\displaystyle \frac{mb}{a-b}}+{\displaystyle \frac{nc}{c-a}}{\textstyle )}=$$

$$={\textstyle (}{\displaystyle \frac{mb-nc}{b-c}}{{\textstyle )}}^{2}+(m-n{)}^{2}{\textstyle (}{\displaystyle \frac{{b}^{2}}{(a-b{)}^{2}}}+{\displaystyle \frac{{c}^{2}}{(c-a{)}^{2}}}{\textstyle )}+2(m-n){\displaystyle \frac{bc(m-n)-a(mb-nc)}{(a-b)(c-a)}}=$$

$$={\textstyle (}{\displaystyle \frac{mb-nc}{b-c}}{{\textstyle )}}^{2}+(m-n{)}^{2}{\textstyle (}{\displaystyle \frac{{b}^{2}}{(a-b{)}^{2}}}+{\displaystyle \frac{{c}^{2}}{(c-a{)}^{2}}}+{\displaystyle \frac{2bc}{(a-b)(c-a)}}{\textstyle )}-{\displaystyle \frac{2a(mb-nc)(m-n)}{(a-b)(c-a)}}=$$

$$={\textstyle (}{\displaystyle \frac{mb-nc}{b-c}}{{\textstyle )}}^{2}+(m-n{)}^{2}{\textstyle (}{\displaystyle \frac{b}{a-b}}+{\displaystyle \frac{c}{c-a}}{{\textstyle )}}^{2}-{\displaystyle \frac{2a(m-n)(mb-nc)}{(a-b)(c-a)}}=$$

$$(}{\displaystyle \frac{mb-nc}{b-c}}{{\textstyle )}}^{2}+(m-n{)}^{2}{\textstyle (}{\displaystyle \frac{a(c-b)}{(a-b)(c-a)}}{{\textstyle )}}^{2}-{\displaystyle \frac{2a(m-n)(mb-nc)}{(a-b)(c-a)}}=$$

$$={\textstyle (}{\displaystyle \frac{mb-nc}{b-c}}-{\displaystyle \frac{a(m-n)(c-b)}{(a-b)(c-a)}}{{\textstyle )}}^{2}\ge 0.$$

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