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Ta có : \(\left\{{}\begin{matrix}a+bc=a\left(a+b+c\right)+bc=\left(a+b\right)\left(a+c\right)\\b+ca=b\left(a+b+c\right)+ca=\left(b+c\right)\left(a+b\right)\\c+ab=c\left(a+b+c\right)+ab=\left(a+c\right)\left(b+c\right)\end{matrix}\right.\)
Từ đó ta có :
\(P=\Sigma\sqrt{\frac{\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(a+b\right)}{\left(a+c\right)\left(b+c\right)}}\)
\(P=\Sigma\sqrt{\left(a+b\right)^2}\)
\(P=\Sigma\left(a+b\right)\)
\(P=2\left(a+b+c\right)\)
\(P=2\)
Ta có: \(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2\ge2xy\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Rightarrow4.2011a\left(2011a-2\right)\le\left(2011a+2011a-2\right)^2=4\left(2011a-1\right)^2\)
\(\Leftrightarrow2011a\left(2011a-2\right)\le\left(2011a-1\right)^2\)
\(\Leftrightarrow\frac{2011a\left(2011a-2\right)}{\left(2011a-1\right)^2}\le1\)
\(\Leftrightarrow\frac{1}{a}-\frac{2011a\left(2011a-2\right)}{\left(2011a-1\right)^2}\ge\frac{1}{a}-1\)\(\Leftrightarrow\frac{1}{a\left(2011a-1\right)^2}\ge\frac{1}{a}-1\)
Tương tự: \(\frac{1}{b\left(2011b-1\right)^2}\ge\frac{1}{b}-1;\frac{1}{c\left(2011c-1\right)^2}\ge\frac{1}{c}-1\)
\(\Leftrightarrow\frac{1}{a\left(2011a-1\right)^2}+\frac{1}{b\left(2011b-1\right)^2}+\frac{1}{c\left(2011c-1\right)^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-3=2011-3=2008\)
Sai thì thôi nhá bẹn!
\(ab+bc+ca=abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
Đặt \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
\(\frac{a}{bc\left(a+1\right)}=\frac{\frac{1}{x}}{\frac{1}{y}\cdot\frac{1}{z}\left(\frac{1}{x}+1\right)}=\frac{xyz}{x\left(x+1\right)}=\frac{yz}{x+1}\)
Tươn tự rồi cộng vế theo vế:
\(A=\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}\le\frac{\left(x+y\right)^2}{4\left(z+1\right)}+\frac{\left(y+z\right)^2}{4\left(x+1\right)}+\frac{\left(z+x\right)^2}{4\left(y+1\right)}\)
Đặt \(x+y=p;y+z=q;z+x=r\Rightarrow p+q+r=2\)
\(A\le\Sigma\frac{\left(x+y\right)^2}{4\left(z+1\right)}=\Sigma\frac{\left(x+y\right)^2}{4\left[\left(z+y\right)+\left(z+x\right)\right]}=\frac{p^2}{4\left(q+r\right)}+\frac{r^2}{4\left(p+q\right)}+\frac{q^2}{4\left(p+r\right)}\)
Sau khi đổi biến,cô si thì em ra thế này.Ai đó giúp em với :)
\(a+bc=a\left(a+b+c\right)+bc=\left(a+b\right)\left(a+c\right)\)
Tương tự: \(b+ca=\left(a+b\right)\left(b+c\right)\) ; \(c+ab=\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow P=a+b+b+c+c+a=2\left(a+b+c\right)=2\)
Bài 2:b) \(9=\left(\frac{1}{a^3}+1+1\right)+\left(\frac{1}{b^3}+1+1\right)+\left(\frac{1}{c^3}+1+1\right)\)
\(\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\therefore\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le3\)
Ta sẽ chứng minh \(P\le\frac{1}{48}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
Ai có cách hay?
1/Đặt a=1/x,b=1/y,c=1/z ->x+y+z=1.
2a) \(VT=\frac{\left(\frac{1}{a^3}+\frac{1}{b^3}\right)\left(\frac{1}{a}+\frac{1}{b}\right)}{\frac{1}{a}+\frac{1}{b}}\ge\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)^2}{\frac{1}{a}+\frac{1}{b}}\)
\(=\frac{\left[\frac{\left(a^2+b^2\right)^2}{a^4b^4}\right]}{\frac{a+b}{ab}}=\frac{\left(a^2+b^2\right)^2}{a^3b^3\left(a+b\right)}\ge\frac{\left(a+b\right)^3}{4\left(ab\right)^3}\)
\(\ge\frac{\left(a+b\right)^3}{4\left[\frac{\left(a+b\right)^2}{4}\right]^3}=\frac{16}{\left(a+b\right)^3}\)
a+b+c=abc à
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