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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) \(2009^2=\left(2008+1\right)^2=2008^2+2.2008+1\)
\(1+2008^2=2009^2-2.2008=2009^2-2.2009\dfrac{2008}{2009}\)
\(A=\sqrt{2009^2-2.2009.\dfrac{2008}{2009}+\dfrac{2008^2}{2009^2}}+\dfrac{2008}{2009}\)
\(A=\sqrt{\left(2009-\dfrac{2008}{2009}\right)^2}+\dfrac{2008}{2009}=2009-\dfrac{2008}{2009}+\dfrac{2008}{2009}=2009\)
Bài 1:
dự đoán dấu = sẽ là \(a^2=b^2=c^2=\dfrac{1}{2}\) nên cứ thế mà chém thôi .
Ta có: \(\left(a^2+1\right)\left(b^2+1\right)=\left(a^2+\dfrac{1}{2}\right)\left(\dfrac{1}{2}+b^2\right)+\dfrac{1}{2}\left(a^2+b^2\right)+\dfrac{3}{4}\)
Bunyakovsky:\(\left(a^2+\dfrac{1}{2}\right)\left(\dfrac{1}{2}+b^2\right)+\dfrac{1}{2}\left(a^2+b^2\right)+\dfrac{3}{4}\ge\dfrac{1}{2}\left(a+b\right)^2+\dfrac{1}{4}\left(a+b\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\left[\left(a+b\right)^2+1\right]\)
\(VT=\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\dfrac{3}{4}\left[\left(a+b\right)^2+1\right]\left(1+c^2\right)\ge\dfrac{3}{4}\left(a+b+c\right)^2\)(đpcm)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{2}}\)
P/s: còn 1 cách khác nữa đó là khai triển sau đó xài schur . Chi tiết trong tệp BĐT schur .pdf
Bạn ghi nhầm đề bài thì phải, \(a,b,c\in Z\)* mới đúng
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{ac}+\dfrac{2}{bc}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
\(\Leftrightarrow\dfrac{2}{ab}+\dfrac{2}{ac}+\dfrac{2}{bc}=0\Leftrightarrow2\left(\dfrac{a+b+c}{abc}\right)=0\Leftrightarrow a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^3=0\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
\(\Leftrightarrow a^3+b^3+c^3=-3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
Mà \(-3\left(a+b\right)\left(a+c\right)\left(b+c\right)⋮3\Rightarrow a^3+b^3+c^3⋮3\)
b)Áp dụng BĐT AM-GM ta có:
\(\dfrac{\sqrt{a}}{\sqrt{b}}+\dfrac{\sqrt{b}}{\sqrt{a}}\ge2\sqrt{\dfrac{\sqrt{a}}{\sqrt{b}}\cdot\dfrac{\sqrt{b}}{\sqrt{a}}}=2\)
Xảy ra khi \(a=b\)
c)Áp dụng BĐT \(x^2+y^2\ge2xy\) có:
\(VT=\left(\sqrt{a}+\sqrt{b}\right)^2=a+b+2\sqrt{ab}\)
\(\ge2\sqrt{\left(a+b\right)\cdot2\sqrt{ab}}=2\sqrt{2\left(a+b\right)\cdot\sqrt{ab}}=VP\)
Xảy ra khi \(a=b\)
a)\(\dfrac{a^2+3}{\sqrt{a^2+3}}=\sqrt{a^2+3}\ge\sqrt{3}< 2\)\
sai đề
@Akai Haruma
Lời giải:
Áp dụng BĐT AM-GM:
\(\frac{a^2+1}{b}+\frac{b^2+1}{c}+\frac{c^2+1}{a}\geq 3\sqrt[3]{\frac{(a^2+1)(b^2+1)(c^2+1)}{abc}}\geq 3\sqrt[3]{\frac{2\sqrt{a^2}.2\sqrt{b^2}.2\sqrt{c^2}}{abc}}=6(*)\)
Theo BĐT AM-GM:
\(a^2+b^2+c^2\geq ab+bc+ac\)
\(\Rightarrow 3(a^2+b^2+c^2)\geq (a+b+c)^2\)
\(\Rightarrow 9\geq (a+b+c)^2\Rightarrow a+b+c\leq 3\Rightarrow 2(a+b+c)\leq 6(**)\)
Từ \((1);(2)\Rightarrow \frac{a^2+1}{b}+\frac{b^2+1}{c}+\frac{c^2+1}{a}\geq 2(a+b+c)\)
Ta có đpcm.