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Ta có :
+) \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Leftrightarrow ayz+bxz+cxy=0\)
+) \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{zc}\right)=1\)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{ayz+bxz+cxy}{abc}\right)=1\)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(đpcm\right)\)
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Rightarrow\dfrac{abz}{xyz}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\)
\(\Rightarrow\dfrac{abz+bxz+cxy}{xyz}=0\)
\(\Rightarrow abz+bxz+cxy=0\)
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\dfrac{xy}{ab}+2\dfrac{xz}{ac}+2\dfrac{yz}{bc}=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy}{abc}+\dfrac{bxz}{abc}+\dfrac{ayz}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=0\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.\left(\dfrac{0}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.0=1\) \(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\) \(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\) ( đpcm )
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=\dfrac{xbc+yac+zab}{abc}=1\\ \Rightarrow xbc+yac+zab=abc\)
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=\dfrac{ayz+bxz+cxy}{xyz}=0\\ \Rightarrow ayz+bxz+cxy=0\)
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(abc\right)^2}\)
\(\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc+yac+zab\right)^2}\\ =\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2+2abc\left(ayz+bxz+cxy\right)}\)
\(\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2+2abc.0}\\ =\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}=1\)
vậy \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)(đpcm)
\(\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.\dfrac{xyz}{abc}.\left(\dfrac{c}{z}+\dfrac{b}{y}+\dfrac{a}{x}\right)=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.\dfrac{xyz}{abc}.0=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(đpcm\right)\)
Lời giải:
Đặt $\frac{x}{a}=m; \frac{y}{b}=n; \frac{z}{c}=p$ với $m,n,p>0$.
BĐT cần chứng minh tương đương với:
(m^2a+n^2b+p^2c)(a+b+c)\geq (am+bn+cp)^2$
$\Leftrightarrow m^2(ab+ac)+n^2(ba+bc)+p^2(ca+cb)\geq 2abmn+2amcp+2bncp$
$\Leftrightarrow ab(m^2-2mn+n^2)+bc(n^2-2np+p^2)+ca(m^2-2mp+p^2)\geq 0$
$\Leftrightarrow ab(m-n)^2+bc(n-p)^2+ca(m-p)^2\geq 0$
(luôn đúng với $a,b,c>0$)
Ta có đpcm.
* 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
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\)
=> \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Rightarrow ayz+bxz+cxy=0\)
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{0}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
Có:
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Rightarrow ayz+bxz+cxy=0\)
Có:
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1^2\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{ayz+bxz+cxy}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(ayz+bxz+cxy=0\right)\)
Vì \(a,b,c,x,y,z\ne0\) nên :
Đặt \(\dfrac{a}{x}=m;\dfrac{b}{y}=n;\dfrac{c}{z}=p\Rightarrow\dfrac{x}{a}=\dfrac{1}{m};\dfrac{y}{b}=\dfrac{1}{n};\dfrac{z}{c}=\dfrac{1}{p}\)
Vậy ta có: \(m+n+p=0\)
\(\dfrac{1}{m}+\dfrac{1}{n}+\dfrac{1}{p}=1\Leftrightarrow\left(\dfrac{1}{m}+\dfrac{1}{m}+\dfrac{1}{p}\right)^2=1\)
\(\Leftrightarrow\dfrac{1}{m^2}+\dfrac{1}{n^2}+\dfrac{1}{p^2}+2\left(\dfrac{m+n+p}{mnp}\right)=1\)
\(\Leftrightarrow\dfrac{1}{m^2}+\dfrac{1}{n^2}+\dfrac{1}{p^2}=1\)
Vậy: \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\Rightarrow M=1\)
