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

  1. 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. 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. 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. 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. 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).

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