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\(VT\le\dfrac{x}{2x+2y+2}+\dfrac{y}{2yz+2z+2}+\dfrac{z}{2z+2x+2}\)
Nên ta chỉ cần chứng minh: \(\dfrac{x}{x+y+1}+\dfrac{y}{y+z+1}+\dfrac{z}{z+x+1}\le1\)
\(\Leftrightarrow\dfrac{y+1}{x+y+1}+\dfrac{z+1}{y+z+1}+\dfrac{x+1}{z+x+1}\ge2\)
Thật vậy, ta có:
\(VT=\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(z+x+1\right)}+\dfrac{\left(y+1\right)^2}{\left(y+1\right)\left(x+y+1\right)}+\dfrac{\left(z+1\right)^2}{\left(z+1\right)\left(y+z+1\right)}\)
\(VT\ge\dfrac{\left(x+y+z+3\right)^2}{\left(x^2+y^2+z^2\right)+3\left(x+y+z\right)+xy+yz+zx+3}\)
\(VT\ge\dfrac{6\left(x+y+z\right)+2\left(xy+yz+zx\right)+12}{3\left(x+y+z\right)+xy+yz+zx+6}=2\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\Leftrightarrow\dfrac{ayz}{xyz}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\Leftrightarrow ayz+bxz+cxy=0\) (1)
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+ayz+bxz}{abc}\right)=1\)
Kết hợp với (1) ta có đpcm
Trước hết ta chứng minh BĐT sau: \(\dfrac{a^2}{x}+\dfrac{b^2}{y}\ge\dfrac{\left(a+b\right)^2}{x+y}\) (*) với \(a,b,x,y>0\). Thật vậy, (*) tương đương \(\dfrac{a^2y+b^2x}{xy}\ge\dfrac{a^2+2ab+b^2}{x+y}\)
\(\Leftrightarrow a^2xy+a^2y^2+b^2x^2+b^2xy\ge2abxy+a^2xy+b^2xy\)
\(\Leftrightarrow\left(ay-bx\right)^2\ge0\) (luôn đúng)
Vậy BĐT được chứng minh. ĐTXR \(\Leftrightarrow ay=bx\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}\)
Áp dụng BĐT (*) liên tiếp, ta được:
\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\ge\dfrac{\left(a+b\right)^2}{x+y}+\dfrac{c^2}{z}\ge\dfrac{\left(a+b+c\right)^2}{x+y+z}\)
ĐTXR \(\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
Ta có đpcm.
Áp dụng bđt bunhiacopxki có:
\(\left(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\right)\left(x+y+z\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\ge\dfrac{\left(a+b+c\right)^2}{x+y+z}\)
Dấu "=" xảy ra <=> \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
BĐT này gọi là BĐT Cauchy-Schwarz đó bạn.
Chứng minh BĐT: \(\dfrac{a^2}{x}+\dfrac{b^2}{y}\ge\dfrac{\left(a+b\right)^2}{x+y}\)
\(\Rightarrow\dfrac{a^2y+b^2x}{xy}\ge\dfrac{\left(a+b\right)^2}{x+y}\Rightarrow\left(a^2y+b^2x\right)\left(x+y\right)\ge\left(a+b\right)^2.xy\)
\(\Leftrightarrow a^2y^2+b^2x^2-2abxy\ge0\Leftrightarrow\left(ay-by\right)^2\ge0\) (luôn đúng)
Áp dụng BĐT trên vào đề:
Ta được: \(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\ge\dfrac{\left(a+b\right)^2}{x+y}+\dfrac{c^2}{z}\ge\dfrac{\left(a+b+c\right)^2}{x+y+z}\)
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\Leftrightarrow ayz+bxz+cxy=0\left(1\right)\)
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=1=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xyc+ayz+xbz}{abc}\right)=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\)(đpcm)
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\Leftrightarrow ayz+bxz+cxy=0\)
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2-2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ac}\right)\)
\(=\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2-2\left(\dfrac{cxy+ayz+bzx}{abc}\right)\)\(=1-0=1\left(dpcm\right)\)