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Thật sự ra mục đích bài này đi chứng minh biểu thức trong ngoặc là scp
Đây là dề thi HSG toán cấp tỉnh Đồng Tháp
Có: \(\sqrt{\left(1+x^2\right)\left(1+y^2\right)\left(1+z^2\right)}\)
\(=\sqrt{\left(x^2+xy+yz+xz\right)\left(y^2+xy+yz+xz\right)\left(z^2+xy+yz+xz\right)}\)
Sau đó thực hiên phân tích đa thức thành nhân tử mỗi ngoặc
\(=\sqrt{\left(x+y\right)^2\left(y+z\right)^2\left(x+z\right)^2}\)
\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)là số hữu tỉ
Vậy
Câu số 1b đề thi hsg
Chào anh từ huyện Cao Lãnh
Bạn tham khảo tại đây nhé :
https://olm.vn/hoi-dap/tim-kiem?id=663631&subject=1&q=ch%E1%BB%A9ng+minh:1/(x-y)%5E2+1/(y-z)%5E2+1/(z-x)%5E2+l%C3%A0+b%C3%ACnh+ph%C6%B0%C6%A1ng+c%E1%BB%A7a+m%E1%BB%99t+s%E1%BB%91+h%E1%BB%AFu+t%E1%BB%89
\(\left\{{}\begin{matrix}x-y=a\\y-z=b\\z-x=c\end{matrix}\right.\Leftrightarrow a+b+c=0\)
\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{\left(y-z\right)^2}+\dfrac{1}{\left(z-x\right)^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
\(=\dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}=\dfrac{\left(ab+bc+ac\right)^2-2abc\left(a+b+c\right)}{a^2b^2c^2}\)
\(=\left(\dfrac{ab+bc+ac}{abc}\right)^2=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\) là bp 1 số hữu tỉ(đpcm)
BĐT cần chứng minh tương đương
\(VT\ge4\left(x+y+z\right)\)
\(\Leftrightarrow\sum\dfrac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge4\left(x+y+z\right)\)
Theo BĐT Cauchy-Schwarz và AM-GM, ta có:
\(\sum\dfrac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge\dfrac{\left(y+z\right)\left(x+\sqrt{yz}\right)}{x}=y+z+\dfrac{\left(y+z\right)\sqrt{yz}}{x}\ge y+z+\dfrac{2yz}{x}\)
Suy ra: \(\sum\dfrac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge2\left(x+y+z\right)-2\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)\)
Mặt khác, theo AM-GM:
\(\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)^2\ge3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\ge x+y+z\)
\(\Rightarrow\sum\dfrac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge4\left(x+y+z\right)\)
Đẳng thức xảy ra khi và chỉ khi \(x=y=z=\dfrac{\sqrt{2}}{3}\)
@Phương An
Câu 1:
\(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}=\dfrac{1}{2}\left(x+y+z\right)\\ \Leftrightarrow2\sqrt{x-a}+2\sqrt{y-b}+2\sqrt{z-c}=x+y+z\\ \Leftrightarrow x+y+z-2\sqrt{x-a}-2\sqrt{y-b}-2\sqrt{z-c}=0\\ \Leftrightarrow x+y+z-2\sqrt{x-a}-2\sqrt{y-b}-2\sqrt{z-c}+3-a-b-c=0\\ \Leftrightarrow\left[\left(x-a\right)-2\sqrt{x-a}+1\right]+\left[\left(y-b\right)-2\sqrt{y-b}+1\right]+\left[\left(z-c\right)-2\sqrt{z-c}+1\right]=0\\ \Leftrightarrow\left(\sqrt{x-a}-1\right)^2+\left(\sqrt{y-b}-1\right)^2+\left(\sqrt{z-c}-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-a}-1=0\\\sqrt{y-b}-1=0\\\sqrt{z-c}-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-a}=1\\\sqrt{y-b}=1\\\sqrt{z-c}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-a=1\\y-b=1\\z-c=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=a+1\\y=b+1\\z=c+1\end{matrix}\right.\)Vậy \(\left\{x;y;z\right\}=\left\{a+1;b+1;c+1\right\}\)
Câu 2:
\(\text{ a) Ta có }:\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}< \dfrac{2}{\sqrt{n-1}+\sqrt{n}}=\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{\left(\sqrt{n-1}+\sqrt{n}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}\\ =\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{n-n+1}=2\left(\sqrt{n}-\sqrt{n-1}\right)\left(1\right)\)
\(\text{Lại có: }\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}>\dfrac{2}{\sqrt{n+1}+\sqrt{n}}=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\\ =\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{n+1-n}=2\left(\sqrt{n+1}-\sqrt{n}\right)\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow2\left(\sqrt{n+1}-n\right)< \dfrac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
b) Áp dụng bất đảng thức ở câu a:
\(\Rightarrow S=1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}\\ >2\left(\sqrt{101}-\sqrt{100}\right)+...+\left(\sqrt{4}-\sqrt{3}\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{1}\right)\\ =2\left(\sqrt{101}-\sqrt{100}+...+\sqrt{4}-\sqrt{3}+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}\right)\\ =2\left(\sqrt{101}-\sqrt{1}\right)>2\left(\sqrt{100}-1\right)=2\left(10-1\right)=18\left(3\right)\)
\(\Rightarrow S=1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}< 2\left(\sqrt{100}-\sqrt{99}\right)+...+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{1}-\sqrt{0}\right)\\ =2\left(\sqrt{100}-\sqrt{99}+...+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}+\sqrt{1}\right)\\ =2\cdot\sqrt{100}=2\cdot10=20\left(4\right)\)
Từ \(\left(3\right)\) và \(\left(4\right)\Rightarrow18< S< 20\)
\(\dfrac{1}{\left[\left(x+z\right)-\left(y+z\right)\right]^2}+\dfrac{1}{\left(x+z\right)^2}+\dfrac{1}{\left(y+z\right)^2}\ge4\)
\(\Leftrightarrow\dfrac{1}{\left(x+z\right)^2+\left(y+z\right)^2-2}+\dfrac{\left(x+z\right)^2+\left(y+z\right)^2-2}{1}\ge2\)
(AM-GM)
\(\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2=x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)=4\)
mà \(x+y+z=2\Rightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=1\)----->thay vào
Bạn có thể giải rõ ràng hơn được không? Mình cũng tự làm được đến đoạn này rồi nhưng k biết thay ntn?????
Ta có:
\(\left(\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{z-x}\right)^2=\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{\left(y-z\right)^2}+\dfrac{1}{\left(z-x\right)^2}+2\left(\dfrac{x-y+y-z+z-x}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\right)=\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{\left(y-z\right)^2}+\dfrac{1}{\left(z-x\right)^2}\)
Vậy: \(\sqrt{\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{\left(y-z\right)^2}+\dfrac{1}{\left(z-x\right)^2}}=\sqrt{\left(\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{z-x}\right)^2}=\)
$=/$\frac{1}{x-y}+\frac{1}{y-z}+\frac{1}{z-x}$/ ($dpcm$)