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Bổ đề : \(x^3+y^3\ge xy\left(x+y\right)=x^2y+xy^2\)
C/m bổ đề : \(x^3+y^3\ge xy\left(x+y\right)\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\ge0\)
Vậy bổ đề đúng .
Áp dụng vào bài toán
\(\dfrac{1}{x^3+y^3+1}+\dfrac{1}{y^3+z^3+1}+\dfrac{1}{z^3+x^3+1}\le1\)
Ta có : \(x^3+y^3+1\ge xy\left(x+y\right)+1=xy\left(x+y\right)+xyz=xy\left(x+y+z\right)\)
\(\Leftrightarrow\dfrac{1}{x^3+y^3+1}\le\dfrac{xyz}{xy\left(x+y+z\right)}=\dfrac{z}{x+y+z}\)
Chứng minh tương tự ta được : \(\dfrac{1}{y^3+z^3+1}\le\dfrac{x}{x+y+z}\)
\(\dfrac{1}{z^3+x^3+1}\le\dfrac{y}{x+y+z}\)
Cộng từng về ta được :
\(\dfrac{1}{x^3+y^3+1}+\dfrac{1}{y^3+z^3+1}+\dfrac{1}{z^3+x^3+1}\ge\dfrac{x+y+z}{x+y+z}=1\)
=> ĐPCM .
a: Thiếu vế phải rồi bạn
b: \(\Leftrightarrow\dfrac{x+y}{xy}>=\dfrac{4}{x+y}\)
\(\Leftrightarrow\left(x+y\right)^2>=4xy\)
\(\Leftrightarrow\left(x-y\right)^2>=0\)(luôn đúng)
* Ta có:
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Leftrightarrow\dfrac{axy}{xyz}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\)
\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Leftrightarrow ayz+bxz+cxy=0\)
* Ta có:
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\dfrac{xy}{ab}+2\dfrac{xz}{ac}+2\dfrac{yz}{bc}=1\)\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{b^2}+2\left(\dfrac{cxy}{abc}+\dfrac{bxz}{abc}+\dfrac{ayz}{abc}\right)=1\)\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=1\)Mà \(cxy+bxz+ayz=0\)
\(\Rightarrow2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=0\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
Vậy.........................
Ta có:
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
=>\(\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
=> \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=1\)
=>\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy}{abc}+\dfrac{ayz}{abc}+\dfrac{bxz}{abc}\right)=1\) (1)
Lại có:
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
=> \(\dfrac{a}{x}.\dfrac{yz}{yz}+\dfrac{b}{y}.\dfrac{xz}{xz}+\dfrac{c}{z}.\dfrac{xy}{xy}=0\)
=>\(\dfrac{ayz}{xuy}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\) (2)
Thay (2) vào (1) ta được
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\)
=> \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
\(\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}+3\)
\(< =>\dfrac{y+z}{x}+1+\dfrac{x+z}{y}+1+\dfrac{x+y}{z}+1\)
\(< =>\dfrac{y+z}{x}+\dfrac{x}{x}+\dfrac{x+z}{y}+\dfrac{y}{y}+\dfrac{x+y}{z}+\dfrac{z}{z}\)
\(< =>\dfrac{x+y+z}{x}+\dfrac{x+y+z}{y}+\dfrac{x+y+z}{z}\) (1)
Thay x+y+z=0vào (1), ta có:\(\dfrac{0}{x}+\dfrac{0}{y}+\dfrac{0}{z}=0+0+0=0\)