# Các phép đặt hay dùng trong chứng minh bất đẳng thức

Tôi sẽ post vài phép đổi biến hay dùng trong topic này.

## 8 thoughts on “Các phép đặt hay dùng trong chứng minh bất đẳng thức”

1. trungtuan says:

Dinh li 1. Cho cac so thuc duong $a,b,c$. Khi do $abc=1$ khi va chi khi ton tai cac so thuc duong $x,y,z$ sao cho $a=\dfrac{x}{y},b=\dfrac{y}{z},c=\dfrac{z}{x}$.

Chung minh. Dieu kien du la hien nhien, voi dieu kien can chung ta chi can chon $x=1,y=\dfrac{1}{a},z=c$.$\Box$

Sau day la cac bai tap ap dung:

Bai 1.1(IMO).
Cho cac so thuc duong $a,b,c$ thoa man $abc=1$. Chung minh rang $(a-1+\dfrac{1}{b})(b-1+\dfrac{1}{c})(c-1+\dfrac{1}{a})\leq 1.$

Bai 1.2.
Chung minh rang voi moi ba so thuc duong $a,b,c$ ta co $\dfrac{a}{a+2b}+\dfrac{b}{b+2c}+\dfrac{c}{c+2a}\geq 1.$

Bai 1.3(Crux).
Cho cac so thuc duong $a,b,c,d,e$ thoa man $abcde=1$. Chung minh rang $\dfrac{a+abc}{1+ab+abcd}+\dfrac{b+bcd}{1+bc+bcde}+\dfrac{c+cde}{1+cd+cdea}+\\ \dfrac{d+dea}{1+de+deab}+\dfrac{e+eab}{1+ea+eabc}\geq \dfrac{10}{3}.$

2. Tuan says:

Dinh li 2. Cho cac so thuc duong $x,y,z$. Khi do $xy+yz+zx+2xyz=1$ khi va chi khi ton tai cac so thuc duong $a,b,c$ sao cho $x=\dfrac{a}{b+c},y=\dfrac{b}{c+a},z=\dfrac{c}{a+b}$.

Chung minh. Dieu kien du la hien nhien, chung ta co the kiem tra truc tiep dang thuc $xy+yz+zx+2xyz=1$. Voi dieu kien can chung ta chi can chon $a=1,b=\dfrac{y(1+x)}{x(1+y)},c=\dfrac{1-xy}{x(1+y)}$. Luu y rang tu dang thuc $xy+yz+zx+2xyz=1$ ta co $1>xy$ va do do $c>0. \Box$

Sau day la cac bai tap ap dung:

Bai 2.1.
Cho cac so thuc duong $a,b,c$. Chung minh rang $\sum_{\text{cyclic}}\dfrac{a}{b+c}\geq \dfrac{3}{2}.$
Bai 2.2.
Cho cac so thuc duong $x,y,z$ thoa man $xy+yz+zx+2xyz=1$. Chung minh rang
a)$xyz\leq\dfrac{1}{8}$.
b)$xy+yz+zx\geq\dfrac{3}{4}$.
c)$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\geq 4(x+y+z).$
Bai 2.3.
Cho cac so thuc khong am $x,y,z$ thoa man $xy+yz+zx+xyz=4$. Chung minh rang $x+y+z\geq xy+yz+zx.$

3. Phan Sy Quang says:

Bai 2.2a, Dat $\sqrt[3]{xyz}=t$
Theo AM-GM ta co :
$ab+bc+ca\ge 3.t^2$
Cho nen:$2t^3+3t^2\le 1$
Hay :$(2t-1)(t+1)^2\le 0$
Suy ra $t\le \dfrac{1}{2}$ hay $abc\le \dfrac{1}{8}$

b,Vi $abc\le \dfrac{1}{8}$ nen $ab+bc+ca\ge \dfrac{3}{4}$

c,Theo bo de tren ton tai $a,b,c$ sao cho:
$x=\dfrac{a}{b+c},y=\dfrac{b}{c+a},z=\dfrac{c}{a+b}$
Bat dang thuc can chung minh tuong duong voi :
$\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\ge 4.(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b})$(1)
den day xai SS cai.
Bien doi(1) tuong duong voi:
$M(a-b)^2+N(a-c)(b-c)\ge 0$
Trong do :
$M=\dfrac{2}{ab}-\dfrac{1}{(a+c)(b+c)}$
$N=\dfrac{1}{ac}+\dfrac{1}{bc}-\dfrac{2(a+b+2c)}{(a+b)(b+c)(c+a)}$
Vi $\dfrac{1}{ab}>\dfrac{1}{(a+c)(b+c)}$nen $M>0$
Ta se chung minh $N>0$
That vay ta co $\dfrac{1}{ac}+\dfrac{1}{bc}\ge \dfrac{4}{c(a+b)}$
Do do ta se chung minh:$\dfrac{4}{c(a+b)}\ge \dfrac{2(a+b+2c)}{(a+b)(b+c)(c+a)}$
Hay $2(a+c)(b+c)>c(a+b+2c)\leftrightarrow2ab+ac+bc>0$
hien nhien.
Bat dang thuc duoc chung minh xong.

4. Phan Sy Quang says:

Bai 2.3
Dat $xyz=d_3,xy+yz+zx=\dfrac{d_2}{3},x+y+z=\dfrac{d_1}{3}$
Tu gia thiet suy ra $3d_2+d_3=4$
Theo BDT Schur $3.d_1^3+d_3\ge 4d_1.d_2.$
$\rightarrow 3d_1^2.d_2^2+d_3.d_1^3\ge 3d_1^2.d_2^2+\dfrac{d_2^2.d_3}{d_1}\ge 4d_2^3$
$\rightarrow (d_1-d_2)^3+3(d_1^3-d_2^3)\ge 0$
Tu do $d_1\ge d_2$ hay $x+y+z\ge xy+yz+zx$.
Bai nay chac dung dat nhu tren cung duoc nhung ,giai the nay ngan gon hon.:D
Quang

5. trungtuan says:

Dinh li 3. Cho cac so thuc duong $x,y,z$. Khi do $xyz=x+y+z+2$(*) khi va chi khi ton tai cac so thuc duong $a,b,c$ sao cho $x=\dfrac{b+c}{a},y=\dfrac{c+a}{b},z=\dfrac{a+b}{c}$.
Chung minh. Viet lai dang thuc (*) duoi dang $1=\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}+\dfrac{2}{xyz}$ va de y toi Dinh li 2 chung ta co dieu phai chung minh.$\Box$

Sau day la cac bai tap ap dung:

Bai 3.1.
Chung minh rang neu $x,y,z>0$ va $xyz=x+y+z+2$ thi
a)$2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})\leq x+y+z+6$.
b)$xyz(x-1)(y-1)(z-1)\leq 8.$
c)$xy+yz+zx\geq 2(x+y+z)$.
d)$\sqrt{x}+\sqrt{y}+\sqrt{z}\leq \dfrac{3}{2}\sqrt{xyz}.$
Bai 3.2(USA MO 2003).
Chung minh rang $\sum_{\text{cyclic}}\dfrac{(2a+b+c)^2}{2a^2+(b+c)^2}\leq 8\forall a,b,c>0.$
Bai 3.3.
Cho cac so thuc duong $x,y,z$ thoa man $xy+yz+zx=2(x+y+z)$. Chung minh rang $xyz\leq x+y+z+2$.
Bai 3.4.
Chung minh rang neu $a,b,c>0$ va $x=a+\dfrac{1}{b},y=b+\dfrac{1}{c},z=c+\dfrac{1}{a}.$ Chung minh rang $xy+yz+zx\geq 2(x+y+z)$.

6. trungtuan says:

Bai 2.2.c sau khi dat nhu the thi Quang chi viec dung cac bat dang thuc $\dfrac{a}{b+c}\leq \dfrac{1}{4}\left(\dfrac{a}{b}+\dfrac{a}{c}\right),...$

7. trungtuan says:

Ban vao muc About cua trang nay nhe! Cam on da ghe tham trang cua minh.

8. Let $a,b,c$ are positive reals numbers such that $ab+bc+ca+abc=4$. Prove that:
$3(a^2+b^2+c^2)+1\ge 10$
I think substitution is not the best way for my above inequality 😛