Cho phân số \(\dfrac{a}{b}\) . Chứng minh rằng :
Nếu \(\dfrac{a-x}{b-y}=\dfrac{a}{b}\) thì \(\dfrac{x}{y}=\dfrac{a}{b}\)
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Ta có: \(\frac{a-x}{b-y}=\frac{a}{b}\Rightarrow\left(a-x\right)b=\left(b-y\right)a\)
\(\Rightarrow ab-bx=ab-ay\Rightarrow bx=ay\)
\(\Rightarrow\frac{x}{y}=\frac{a}{b}\left(ĐPCM\right)\)
Đề bài sai
Ví dụ: với \(a=1;b=2;c=3,d=4\) thì \(x=\dfrac{1}{2}\) ; \(y=\dfrac{3}{4}\) ; \(z=\dfrac{2}{3}\)
Khi đó \(x< y\) nhưng \(z< y\)
\(\text{Vì }\dfrac{a}{b}< \dfrac{c}{d}\text{ nên }ad< bc\left(1\right)\)
\(\text{Xét tích}:a\left(b+d\right)=ab+ad\left(2\right)\)
\(b\left(a+c\right)=ba+bc\left(3\right)\)
\(\text{Từ(1);(2);(3)}\Rightarrow a\left(b+d\right)< b\left(a+c\right)\text{ do đó }\dfrac{a}{b}< \dfrac{a+c}{b+d}\left(4\right)\)
\(\text{Tương tự ta có:}\dfrac{a+c}{b+d}< \dfrac{c}{d}\left(5\right)\)
\(\text{Từ (4);(5) ta được }\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)
\(\Rightarrow x< y< z\)
ĐKXĐ: \(\left\{{}\begin{matrix}a\ne0\\b\ne0\\c\ne0\end{matrix}\right.\)Ta có: \(\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)\cdot\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=\left(a^2+b^2+c^2\right)\cdot\left(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\right)\)
\(\Leftrightarrow x^2+y^2+z^2=x^2+\dfrac{x^2\cdot\left(b^2+c^2\right)}{a^2}+y^2+\dfrac{y^2\left(a^2+c^2\right)}{b^2}+z^2+\dfrac{z^2\cdot\left(a^2+b^2\right)}{c^2}\)
\(\Leftrightarrow x^2\cdot\dfrac{b^2+c^2}{a^2}+y^2\cdot\dfrac{a^2+c^2}{b^2}+z^2\cdot\dfrac{a^2+b^2}{c^2}=0\)(1)
Vì (1) luôn không âm mà a,b,c≠0
nên x=y=z=0
⇒\(\dfrac{x^{2019}+y^{2019}+z^{2019}}{a^{2019}+b^{2019}+c^{2019}}=\dfrac{0^{2019}+0^{2019}+0^{2019}}{a^{2019}+b^{2019}+c^{2019}}=0\)
mà \(\dfrac{x^{2019}}{a^{2019}}+\dfrac{y^{2019}}{b^{2019}}+\dfrac{z^{2019}}{c^{2019}}=\dfrac{0^{2019}}{a^{2019}}+\dfrac{0^{2019}}{b^{2019}}+\dfrac{0^{2019}}{c^{2019}}=0\)
nên \(\dfrac{x^{2019}+y^{2019}+z^{2019}}{a^{2019}+b^{2019}+c^{2019}}=\dfrac{x^{2019}}{a^{2019}}+\dfrac{y^{2019}}{b^{2019}}+\dfrac{z^{2019}}{c^{2019}}\)
b/ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\dfrac{a}{d}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
=> \(\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{c+d+b}\right)^3\) (2)Từ (1) và (2)=>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.
a.
\(\dfrac{x}{x+\sqrt{3x+yz}}=\dfrac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}=\dfrac{x}{x+\sqrt{\left(x+y\right)\left(z+x\right)}}\le\dfrac{x}{x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}}\)
\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Tương tự:
\(\dfrac{y}{y+\sqrt{3y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Cộng vế:
\(VT\le\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
b.
\(VP=\dfrac{4\left(a+b+c\right)}{2\sqrt{4a\left(a+3b\right)}+2\sqrt{4b\left(b+3c\right)}+2\sqrt{4c\left(c+3a\right)}}\)
\(VP\ge\dfrac{4\left(a+b+c\right)}{4a+a+3b+4b+b+3c+4c+c+3a}\)
\(VP\ge\dfrac{4\left(a+b+c\right)}{8\left(a+b+c\right)}=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
\(\dfrac{a-x}{b-y}=\dfrac{a}{b}\)
\(\Rightarrow\dfrac{a-x}{a}=\dfrac{b-y}{b}\)
\(\Rightarrow1-\dfrac{x}{a}=1-\dfrac{y}{b}\)
\(\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{a}{b}\)