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Ta có :
\(\frac{1+\sqrt{1+x^2}}{x}=\frac{2+\sqrt{4\left(1+x^2\right)}}{2x}\le\frac{2+\frac{4+1+x^2}{2}}{2x}=\frac{9+x^2}{4x}\)
tương tự : \(\frac{1+\sqrt{1+y^2}}{y}\le\frac{9+y^2}{4y}\); \(\frac{1+\sqrt{1+z^2}}{z}\le\frac{9+z^2}{4z}\)
\(\Rightarrow\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le\frac{\left(9+x^2\right)yz+\left(9+y^2\right)xz+\left(9+z^2\right)xy}{4xyz}\)
\(=\frac{9\left(xy+yz+xz\right)+xyz\left(x+y+z\right)}{4xyz}\le\frac{9\frac{\left(x+y+z\right)^2}{3}+\left(xyz\right)^2}{4xyz}=\frac{4\left(xyz\right)^2}{4xyz}=xyz\)
Dấu " = " xảy ra khi x = y = z = \(\sqrt{3}\)
Ta chứng minh
\(a+b\ge\sqrt[3]{ab}\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\)
\(\Leftrightarrow\left(\sqrt[3]{a}-\sqrt[3]{b}\right)^2\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\ge0\)(đúng )
Áp đụng vào bài toán ta được
\(\frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{z+x+1}\)
\(\le\frac{1}{\sqrt[3]{xy}\left(\sqrt[3]{x}+\sqrt[3]{y}\right)+1}+\frac{1}{\sqrt[3]{yz}\left(\sqrt[3]{y}+\sqrt[3]{z}\right)+1}+\frac{1}{\sqrt[3]{zx}\left(\sqrt[3]{z}+\sqrt[3]{x}\right)+1}\)
\(=\frac{\sqrt[3]{z}}{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}+\frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}+\frac{\sqrt[3]{y}}{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}=1\)
Ta có: \(xyz=1\)=>\(xy=\frac{1}{z}\)
Theo BĐT cosy, ta có: \(x+y+1\ge3\sqrt[3]{xy}=3\sqrt[3]{\frac{1}{z}}=\frac{3}{3\sqrt[3]{z}}\)
tương tự:\(y+z+1\ge3\sqrt[3]{\frac{1}{x}}=\frac{3}{\sqrt[3]{x}}\)
\(z+x+1\ge3\sqrt[3]{\frac{1}{y}}=\frac{3}{\sqrt[3]{y}}\)
=> \(Q\le\frac{1}{\frac{3}{\sqrt[3]{z}}}+\frac{1}{\frac{3}{\sqrt[3]{x}}}+\frac{1}{\frac{3}{\sqrt[3]{y}}}=\frac{\sqrt[3]{z}}{3}+\frac{\sqrt[3]{x}}{3}+\frac{\sqrt[3]{y}}{3}=\frac{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}{3}\)
Áp dụng BĐT trên lần nữa ta được \(Q\le\frac{3\sqrt[3]{\sqrt[3]{xyz}}}{3}=\frac{3}{3}=1\)
Vậy DTLN của Q=1
dấu "=" xảy ra khi x=y=z=1
Ta có: \(x+y+z=xyz\Rightarrow x=\frac{x+y+z}{yz}\Rightarrow x^2=\frac{x^2+xy+xz}{yz}\Rightarrow x^2+1=\frac{\left(x+y\right)\left(x+z\right)}{yz}\)\(\Rightarrow\sqrt{x^2+1}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{yz}}\le\frac{\frac{x+y}{y}+\frac{x+z}{z}}{2}=1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)\(\Rightarrow\frac{1+\sqrt{1+x^2}}{x}\le\frac{2+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)}{x}=\frac{2}{x}+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)
Tương tự: \(\frac{1+\sqrt{1+y^2}}{y}\le\frac{2}{y}+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)\); \(\frac{1+\sqrt{1+z^2}}{z}\le\frac{2}{z}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Cộng theo vế ba bất đẳng thức trên, ta được: \(\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3.\frac{xy+yz+zx}{xyz}\)\(\le3.\frac{\frac{\left(x+y+z\right)^2}{3}}{xyz}=\frac{\left(x+y+z\right)^2}{xyz}=\frac{\left(xyz\right)^2}{xyz}=xyz\)
Đẳng thức xảy ra khi \(x=y=z=\sqrt{3}\)
Với 2 số dương bất kì: ( 1 )
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)Vì x và y dương nên \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow\left(x-y\right)^2\ge0\forall x;y\)
Áp dụng ( 1 ): \(\frac{4}{2x+y+z}=\frac{4}{\left(x+y\right)+\left(x+z\right)}\le\frac{1}{x+y}+\frac{1}{x+z}\)
Mà: \(\frac{1}{x+y}+\frac{1}{x+z}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}+\frac{1}{z}\right)=\frac{1}{4}\)\(=\frac{1}{4}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Nên: \(\frac{1}{2x+y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Tương tự ta có: \(\frac{1}{x+2y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{2}{y}+\frac{1}{z}\right)\)
Và \(\frac{1}{x+y+2z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Cộng vế với vế các bất đẳng thức kết hợp với điều kiện \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\) nên ta có đpcm
\(A=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\).Áp dụng BĐT Cauchy-Schwarz,ta có:
\(=\left(1-\frac{1}{x+1}\right)+\left(1-\frac{1}{y+1}\right)+\left(1-\frac{1}{z+1}\right)\)
\(=\left(1+1+1\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(\ge3-\frac{9}{\left(x+y+z\right)+\left(1+1+1\right)}=\frac{3}{4}\)
Dấu "=" xảy ra khi x = y = z = 1/3
Vậy A min = 3/4 khi x=y=z=1/3
ui, đề thi HSG huyện mình nè. cậu huyện nào mà đăng thế
chứng minh BĐT : \(a^3+b^3+1\ge ab\left(a+b\right)\) với a>0,b>0
\(\Rightarrow a^3+b^3+1\ge ab\left(a+b\right)+abc=ab\left(a+b+c\right)\)
áp dụng BĐT trên,ta có:
\(x+y+1\ge\sqrt[3]{xy}\left(\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}\right)\)
\(\Rightarrow\frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{x+z+1}\le\frac{1}{\sqrt[3]{xy}\left(\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}\right)}+\frac{1}{\sqrt[3]{yz}\left(\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}\right)}+\frac{1}{\sqrt[3]{xz}\left(\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}\right)}\)
\(=\frac{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}{\sqrt[3]{xyz}\left(\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}\right)}=1\)
Dấu " = " xảy ra khi x = y = z = 1
Ap dung bdt \(a+b\ge\sqrt[3]{a^2b}+\sqrt[3]{ab^2}\left(a,b\ge0\right)\)
ta co \(x+y\ge\sqrt[3]{xy}\left(\sqrt[3]{x}+\sqrt[3]{y}\right)\)
ma \(xyz=1=>\sqrt[3]{xy}=\frac{1}{\sqrt[3]{z}}\)
nen \(x+y\ge\frac{\sqrt[3]{x}+\sqrt[3]{y}}{\sqrt[3]{z}}\)
=> \(x+y+1\ge\frac{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}{\sqrt[3]{z}}\)
=>\(\frac{1}{x+y+1}\le\frac{\sqrt[3]{z}}{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}\)
chung minh tuong tu cung co \(\frac{1}{x+z+1}\le\frac{\sqrt[3]{y}}{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}\) va \(\frac{1}{z+y+1}\le\frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}\)
cong 3 bdt cung chieu ta duoc
\(\frac{1}{x+y+1}+\frac{1}{x+z+1}+\frac{1}{y+z+1}\le\frac{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}=1\)
dau = xay ra khi x=y=z=1
Chuc ban hoc tot !!!