22 thoughts on “Có bao nhiêu cách chứng minh cho bất đẳng thức Nesbit?”

  1. Cach 3.
    Khong giam tong quat ta co the gia su a+b+c=1. Xet ham so f(x)=\dfrac{x}{1-x} tren I=(0,1), ta co f''(x)=-\dfrac{2}{(x-1)^3}>0\forall x\in I. Do do f la ham so loi tren I, theo bat dang thuc Jensen ta co f(a)+f(b)+f(c)\geq 3f(\dfrac{a+b+c}{3})=3f(\dfrac{1}{3})=\dfrac{3}{2}. Nhu vay bat dang thuc duoc chung minh.\Box

  2. Cach 4.
    Dat f(a,b,c)=\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}
    Ta se chung minh:
    f(a,b,c)\ge f(\dfrac{a+b}{2},\dfrac{a+b}{2},c)
    That vay :
    f(a,b,c)- f(\dfrac{a+b}{2},\dfrac{a+b}{2},c)=\dfrac{(a-b)^2.(a+b+c)}{(2c+a+b)(b+c)(a+c)}\geq 0
    Vay ta co f(a,b,c)\ge f(\dfrac{a+b}{2},\dfrac{a+b}{2},c)
    Tiep theo ta chung minh f(\dfrac{a+b}{2},\dfrac{a+b}{2},c)\ge\dfrac{3}{2}
    That vay :Dat t=\dfrac{a+b}{2}
    Bat dang thuc can chung minh tuong duong voi
    \dfrac{2t}{c+t}+\dfrac{c}{2t}\ge \dfrac{3}{2}
    Hay \dfrac{(t-c)^2}{2t.(c+t)}\ge 0
    Hien nhien dung.

  3. Cach 8.
    Bat dang thuc tuong duong voi
    \dfrac{\sum_{cyclic}a^3+[\sum_{sym}a^2.b]+3abc}{[\sum_{sym}a^2.b]+2abc}\ge \dfrac{3}{2}
    tuong duong voi :
    2.\sum_{cyclic}a^3\ge \sum_{sym}a^2.b
    Theo AM-GM:
    a^2.b\le \dfrac{a^3+a^3+b^3}{3}
    Cong ve theo ve 6 BDT tuong tu ta co dpcm.

  4. Cach 9.
    Khong giam tong quat, gia su a\geq b\geq c khi do b+c\leq c+a\leq a+b hay \dfrac{1}{b+c}\geq\dfrac{1}{c+a}\geq \dfrac{1}{a+b} (chu y la a>0,b>0,c>0) va boi bat dang thuc hoan vi ta co
    \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq \dfrac{b}{b+c}+\dfrac{c}{c+a}+\dfrac{a}{a+b}(1) va \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq \dfrac{c}{b+c}+\dfrac{a}{c+a}+\dfrac{b}{a+b}(2) .

    Cong theo ve (1) voi (2) ta co dieu phai chung minh. \Box.

  5. Cach 11.
    Khong giam tong quat , gia su a+b+c=3. Bang bien doi tuong duong ta chung minh duoc bat dang thuc sau \dfrac{x}{3-x}\geq\dfrac{3}{4}(x-1)+\dfrac{1}{2}(0<x<3)(*). Lan luot thay x boi a,b,c trong (*) chung ta duoc ba bat dang thuc, cong theo ve ba bat dang thuc nay chung ta duoc dieu phai chung minh.\Box

  6. Cach 14.
    Ta chung minh nhan xet sau
    \dfrac{a}{b+c}\geq \dfrac{8a-b-c}{4(a+b+c)}

    That vay,no tuong duong voi:(2a-b-c)^2\ge 0
    hien nhien dung.Cong ve theo ve cac bat dang thuc tuong tu ta co ngay dpcm.

  7. Cach 18(Thu luom tren Internet).
    Dat x=\dfrac{a}{b+c},y=\dfrac{b}{c+a},z=\dfrac{c}{a+b} va xet ham so f(t)=\dfrac{t}{t+1}(t>0). Ta co f'(t)=\dfrac{1}{(t+1)^2}>0\forall t>0 va f''(t)=-\dfrac{2}{(t+1)^3}<0\forall t\in (0,\infty). Theo bat dang thuc Jensen ta co f(\dfrac{1}{2})=\dfrac{1}{3}=\dfrac{f(x)+f(y)+f(z)}{3}\leq f\left(\dfrac{x+y+z}{3}\right). Nhung vi f la tang ngat tren (0,\infty) nen \dfrac{1}{2}\leq \dfrac{x+y+z}{3}. Suy ra dieu phai chung minh.\Box

  8. Cach 19.(Tren internet)
    Se la du neu ta chung minh duoc \dfrac{a}{b+c}\geq \dfrac{3a^{3/2}}{2(a^{3/2}+b^{3/2}+c^{3/2})} hay 2(a^{3/2}+b^{3/2}+c^{3/2})\geq 3a^{1/2}(b+c)(*).

    Theo AM-GM ta co
    a^{3/2}+b^{3/2}+b^{3/2}\geq 3a^{1/2}b va
    a^{3/2}+c^{3/2}+c^{3/2}\geq 3a^{1/2}c .
    Cong theo ve hai bat dang thuc cuoi nay ta duoc (*).\Box

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