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Ta có:
\(x+y+z=\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\).
\(x+y+z=\frac{\left(a-b\right)\left(b+c\right)\left(c+a\right)+\left(b-c\right)\left(a+b\right)\left(c+a\right)+\left(c-a\right)\left(a+b\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Ta có:
\(\left(a-b\right)\left(b+c\right)\left(c+a\right)+\left(b-c\right)\left(c+a\right)\left(a+b\right)+\left(c-a\right)\left(a+b\right)\left(b+c\right)\).
\(=\left(c+a\right)\left[\left(a-b\right)\left(b+c\right)+\left(b-c\right)\left(a+b\right)\right]+\left(c-a\right)\left(a+b\right)\left(b+c\right)\).
\(=\left(c+a\right)\left(ab+ac-b^2-bc+ab+b^2-ac-bc\right)\)\(+\left(c-a\right)\left(ab+ac+b^2+bc\right)\).
\(=\left(c+a\right)\left(2ab-2bc\right)-\left(a-c\right)\left(ab+ac+b^2+bc\right)\).
\(=2b\left(c+a\right)\left(a-c\right)-\left(a-c\right)\left(ab+ac+b^2+bc\right)\).
\(=\left(2bc+2ab\right)\left(a-c\right)-\left(a-c\right)\left(ab+ac+b^2+bc\right)\).
\(=\left(a-c\right)\left(2ab+2bc-ab-ac-b^2-bc\right)\).
\(=\left(a-c\right)\left(ab+bc-b^2-ac\right)=\left(a-c\right)\left[\left(ab-b^2\right)-\left(ac-bc\right)\right]\).
\(=\left(a-c\right)\left[b\left(a-b\right)-c\left(a-b\right)\right]=\left(a-c\right)\left(a-b\right)\left(b-c\right)\).
Do đó\(x+y+z=\frac{\left(a-c\right)\left(a-b\right)\left(b-c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\).
Mà \(xyz=\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)nên:
\(x+y+z=-xyz\).
\(\Rightarrow x+y+z+xyz=0\)(điều phải chứng minh).
a ) Đặt A = \(\frac{-a+b+c}{2a}+\frac{a-b+c}{2b}+\frac{a+b-c}{2c}=\frac{1}{2}\left(-1+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}-1+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}-1\right)\)
\(=\frac{1}{2}\left(\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}-3\right)\)
Do a ; b ; c > 0 , áp dụng BĐT Cô - si cho các cặp số dương , ta có :
\(A\ge\frac{1}{2}\left[2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{\frac{b}{c}.\frac{c}{b}}+2\sqrt{\frac{a}{c}.\frac{c}{a}}-3\right]=\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
b ) \(P=\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\frac{x^2}{xy+xz}+\frac{y^2}{xy+yz}+\frac{z^2}{xz+yz}\ge\frac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\frac{3\left(xy+yz+xz\right)}{2\left(xy+yz+xz\right)}=\frac{3}{2}\)
( áp dụng BĐT Cauchy - Schwarz )
Dấu " = " xảy ra \(\Leftrightarrow x=y=z\)
a) Chứng minh được BĐT \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)(*)
Dấu "=" xảy ra <=> a=b
Áp dụng BĐT (*) vào bài toán ta có:
\(\hept{\begin{cases}\frac{1}{2x+y+z}=\frac{1}{x+y+x+y}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\\\frac{1}{x+2y+z}=\frac{1}{x+y+y+z}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\\\frac{1}{x+y+2z}=\frac{1}{x+y+z+z}\le\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\end{cases}}\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\cdot2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
Tiếp tục áp dụng BĐT (*) ta có:
\(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right);\frac{1}{y+z}\le\frac{1}{4}\left(\frac{1}{y}+\frac{1}{z}\right);\frac{1}{z+x}\le\frac{1}{4}\left(\frac{1}{z}+\frac{1}{x}\right)\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\cdot2\cdot\frac{1}{4}\cdot2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)
\(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\)
Dấu "=" xảy ra <=> \(x=y=z=\frac{3}{4}\)
b) áp dụng bđt \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)ta có:
\(\hept{\begin{cases}\frac{1}{a+b-c}+\frac{1}{b+c-a}\ge\frac{4}{a+b-c+b+c-a}=\frac{4}{2b}=\frac{2}{b}\\\frac{1}{b+c-a}+\frac{1}{a+c-b}\ge\frac{4}{b+c-a+a+c-b}=\frac{4}{2c}=\frac{2}{c}\\\frac{1}{a+b-c}+\frac{1}{a+c-b}\ge\frac{4}{a+b-c+a+c-b}=\frac{4}{2a}=\frac{2}{a}\end{cases}}\)
Cộng theo vế 3 BĐT ta có:
\(2VT\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2VP\)
\(\Rightarrow VT\ge VP\)
Đẳng thức xảy ra <=> a=b=c
chịu khó lắm
Ok
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\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\)
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow\frac{xbc+yac+zab}{abc}=1\)
\(\Rightarrow xbc+yac+zab=abc\)
\(\Rightarrow\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2+2.xbc.yac+2.yac.zab+2.xbc.zab=\left(abc\right)^2\)
\(\Rightarrow x^2b^2c^2+y^2a^2c^2+z^2a^2b^2+2abc\left(cxy+ayz+bxz\right)=\left(abc\right)^2\)
\(\Rightarrow x^2b^2c^2+y^2a^2c^2+z^2a^2b^2=a^2b^2c^2\)
\(\Rightarrow\frac{x^2b^2c^2+y^2a^2c^2+z^2a^2b^2}{a^2b^2c^2}=1\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)
\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow ayz+bxz+cxy=0\)
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Leftrightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1-2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)\)
\(=1-2.\frac{cxy+bxz+ayz}{abc}=1-2.0=1\)
Lời giải:
Nếu $x=0$ thì $a=b$. Khi đó:
$x+y+z+xyz=y+z=\frac{b-c}{b+c}+\frac{c-b}{c+b}=0$ (đpcm)
Tương tự: $y=0; z=0$ cũng vậy.
Nếu $xyz\neq 0$:
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=\frac{(a+b)(b+c)}{(a-b)(b-c)}+\frac{(b+c)(c+a)}{(b-c)(c-a)}+\frac{(a+b)(a+c)}{(a-b)(c-a)}\)
\(=\frac{(a+b)(b+c)(c-a)+(b+c)(c+a)(a-b)+(a+b)(a+c)(b-c)}{(a-b)(b-c)(c-a)}\)
\(=\frac{(ab+bc+ac)[(c-a)+(b-c)+(a-b)]+b^2(c-a)+c^2(a-b)+a^2(b-c)}{(a-b)(b-c)(c-a)}\)
\(=\frac{b^2(c-a)+c^2(a-b)+a^2(b-c)}{(a-b)(b-c)(c-a)}=\frac{(a^2b+b^2c+c^2a)-(ab^2+bc^2+ca^2)}{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}=-1\)
\(\Rightarrow \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}+1=0\Leftrightarrow \frac{x+y+z+xyz}{xyz}=0\Rightarrow x+y+z+xyz=0\)
Ta có đpcm.