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\(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(\Rightarrow\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\le\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=2\)
Lại có \(\dfrac{1}{2x+y+z}=\dfrac{1}{x+y+x+z}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\)
Tương tự \(\dfrac{1}{x+2y+z}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}\right)\)
\(\dfrac{1}{x+y+2z}\le\dfrac{1}{4}\left(\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)\)
Cộng vế với vế: \(P\le\dfrac{1}{2}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)=\dfrac{1}{2}.2=1\)
\(\Rightarrow P_{max}=1\) khi \(x=y=z=\dfrac{3}{4}\)
c.
\(\left(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right)^2=2010\)
\(\leftrightarrow\) \(x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+1+x^2+y^2+x^2y^2=2010\)
\(\leftrightarrow\)\(x^2+x^2y^2+2x\sqrt{1+y^2}.y\sqrt{1+x^2}+y^2+x^2y^2=2009\)
\(\leftrightarrow\) \(\left(x\sqrt{1+y^2}+y\sqrt{1+x^2}\right)^2=2009\)
\(\leftrightarrow\) \(x\sqrt{1+y^2}+y\sqrt{1+x^2}=\sqrt{2009}\)
c) \(A^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=x^2y^2+x^2+x^2y^2+y^2+1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-1\)
\(=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-1\)
\(=\left[xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right]^2-1=2010-1=2009\)
Vì A>0 nên \(A=\sqrt{2009}\)
d) \(2009^2=\left(2008+1\right)^2=2008^2+2.2008+1\)
\(1+2008^2=2009^2-2.2008=2009^2-2.2009\dfrac{2008}{2009}\)
\(A=\sqrt{2009^2-2.2009.\dfrac{2008}{2009}+\dfrac{2008^2}{2009^2}}+\dfrac{2008}{2009}\)
\(A=\sqrt{\left(2009-\dfrac{2008}{2009}\right)^2}+\dfrac{2008}{2009}=2009-\dfrac{2008}{2009}+\dfrac{2008}{2009}=2009\)
Áp dụng liên tiếp bất đẳng thức Mincopxki và bất đẳng thức Cauchy-Schwarz:
\(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(A\ge\sqrt{4+\dfrac{81}{4}}=\sqrt{\dfrac{97}{4}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
\(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B=\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\dfrac{162}{\left(x+y+z\right)^2}}\)
\(B\ge\sqrt{4+\dfrac{162}{4}}=\sqrt{\dfrac{89}{2}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
Bài 5: Đặt \(t=\dfrac{\left(x+y+1\right)^2}{xy+x+y}\)
Ta đã biết bđt quen thuộc là \(x^2+y^2+1\ge xy+x+y\)
Vậy nên ta sẽ chứng minh \(t\geq 3\)
Thật vậy: \(t\geq 3\Leftrightarrow 2(x+y+1)^2\geq 6(x+y+xy)\)
\(\Leftrightarrow (x-y)^2+(x-1)^2+(y-1)^2\geq 0\)
Đẳng thức xảy ra khi và chỉ khi \(x=y=1\)
Ta có: \(A=\dfrac{8t}{9}+\left(\dfrac{t}{9}+\dfrac{1}{t}\right)\geq \dfrac{24}{9}+\dfrac{2}{3}=\dfrac{10}{3}\)
Dấu "=" xảy ra khi \(t=3\Leftrightarrow x=y=1\)
3)
x^2 = 2x + \(\sqrt{2x-1}\) \(\Rightarrow\) x^2 = ( 2x -1 ) + \(\sqrt{2x-1}\) +1
\(\Rightarrow\) x^2 = (\(\sqrt{2x-1}\) + 1)^2 chuyển vế rồi phân tích thành nhân tử là ok