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Giả sử \(A=B\)\(\Leftrightarrow\)\(A-B=0\)\(\Leftrightarrow\)\(\frac{x^2}{x+y}-\frac{y^2}{x+y}+\frac{y^2}{y+z}-\frac{z^2}{y+z}+\frac{z^2}{z+x}-\frac{x^2}{z+x}=0\)
\(\Leftrightarrow\)\(\frac{\left(x-y\right)\left(x+y\right)}{x+y}+\frac{\left(y-z\right)\left(y+z\right)}{y+z}+\frac{\left(z-x\right)\left(z+x\right)}{z+x}=0\)
\(\Leftrightarrow\)\(x-y+y-z+z-x=0\)
\(\Leftrightarrow\)\(0=0\) ( đúng )
Vậy điều giả sử đúng hay \(A=B\)
...
Ta có \(VT=\frac{x}{1-x^2}+\frac{y}{1-y^2}+\frac{z}{1-z^2}\)
Lại có \(x^2\left(1-x^2\right)^2=\frac{2x^2\left(1-x^2\right)\left(1-x^2\right)}{2}\le\frac{\left(2x^2+1-x^2+1-x^2\right)^3}{54}=\frac{4}{27}\)
\(\Leftrightarrow\) \(x\left(1-x^2\right)\le\frac{2}{3\sqrt{3}}\) \(\Leftrightarrow\) \(\frac{1}{x\left(1-x^2\right)}\ge\frac{3\sqrt{3}}{2}\) \(\Leftrightarrow\) \(\frac{x}{\left(1-x^2\right)}\ge\frac{3\sqrt{3}}{2}x^2\) (1)
Tương tự cho \(\frac{y}{\left(1-y^2\right)}\ge\frac{3\sqrt{3}}{2}y^2\) (2) và \(\frac{z}{\left(1-z^2\right)}\ge\frac{3\sqrt{3}}{2}z^2\) (3)
Cộng vế theo vế ta được \(VT=\frac{x}{1-x^2}+\frac{y}{1-y^2}+\frac{z}{1-z^2}\ge\frac{3\sqrt{3}}{2}\left(x^2+y^2+z^2\right)=\frac{3\sqrt{3}}{2}\)
Đẳng thức xảy ra khi và chỉ khi \(x=y=z=\frac{\sqrt{3}}{3}\)
Từ \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)
\(\Rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1^2\)
\(\left(\frac{x}{a}+\frac{y}{b}\right)^2+2\left(\frac{x}{a}+\frac{y}{b}\right)\frac{z}{c}+\left(\frac{z}{c}\right)^2=1\)
\(\left(\frac{x}{a}\right)^2+2\frac{x}{a}\frac{y}{b}+\left(\frac{y}{b}\right)^2+\left(2\frac{x}{a}+2\frac{y}{b}\right)\frac{z}{c}+\left(\frac{z}{c}\right)^2=1\)
\(\frac{x^2}{a^2}+\frac{2xy}{ab}+\frac{y^2}{b^2}+\frac{2xz}{ac}+\frac{2yz}{bc}+\frac{z^2}{c^2}=1\)
\(\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)+\left(\frac{2xy}{ab}+\frac{2xz}{ac}+\frac{2yz}{bc}\right)=1\)
\(\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)+\frac{2xyz}{abc}\left(\frac{c}{z}+\frac{b}{y}+\frac{a}{x}\right)=1\)
\(\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)+\frac{2xyz}{abc}.0=1\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(ĐPCM\right)\)
\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow ayz+bxz+cxy=0\)
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Leftrightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1-2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)\)
\(=1-2.\frac{cxy+bxz+ayz}{abc}=1-2.0=1\)
a) \(A=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}+\frac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{2\left(y-z\right)\left(z-x\right)+2\left(x-y\right)\left(z-x\right)+2\left(x-y\right)\left(y-z\right)+\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left[\left(x-y\right)+\left(y-z\right)+\left(z-x\right)\right]^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y+y-z+z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)
Áp dụng: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
b)Ta có: \(\frac{x^2}{y+z}+x=\frac{x^2+x\left(y+z\right)}{y+z}=\frac{x^2+xy+xz}{y+z}=\frac{x\left(x+y+z\right)}{y+z}\)
Tương tự: \(\frac{y^2}{x+z}+y=\frac{y^2+xy+zy}{x+z}=\frac{y\left(x+y+z\right)}{x+z}\)
\(\frac{z^2}{x+y}+z=\frac{z^2+xz+zy}{x+y}=\frac{z\left(x+y+z\right)}{x+y}\)
Suy ra: \(A+\left(x+y+z\right)\)
\(=\frac{x\left(x+y+z\right)}{y+z}+\frac{y\left(x+y+z\right)}{z+x}+\frac{z\left(x+y+z\right)}{x+y}+\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}+1\right)\)
\(=2.\left(x+y+z\right)\)
Nên \(A=2.\left(x+y+z\right)-\left(x+y+z\right)=x+y+z\)
Mình có sai chỗ nào không nhỉ?
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1^2\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{cxy+ayz+bxz}{abc}\right)=1\)
Mà \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.\frac{0}{abc}=1\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.0=1\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(dpcm\right)\)
Chúc bạn học tốt
1 cái T I C K nha cảm ơn
Đầu tiên ta sẽ chứng minh \(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{\left(x+y\right)^2}{a+b}\left(1\right)\)
\(\Leftrightarrow x^2b\left(a+b\right)+y^2a\left(a+b\right)\ge ab\left(x+y\right)^2\)
\(\Leftrightarrow\left(bx-ay\right)^2\ge0\left(LĐ\right)\)
Dấu "=" xảy ra khi \(\frac{x}{a}=\frac{y}{b}\)
Vậy BĐT (1) đã được chứng minh
Với 6 số x,y,z,a,b,c >0 ta sẽ áp dụng BĐT (1) hai lần:
\(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y\right)^2}{a+b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}\left(đpcm\right)\)
Bài làm:
Áp dụng Cauchy Schwars ta có:
\(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}\)
Dấu "=" xảy ra khi: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
Ta có:
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)
\(\Leftrightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.\frac{xyz}{abc}\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(đpcm\right)\)
Xét hiệu A - B là ra thôi mà