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\(\dfrac{1}{\sqrt{a^2-ab+b^2}}< =\dfrac{1}{\sqrt{2ab-ab}}=\dfrac{1}{\sqrt{ab}}\)
\(\sqrt{\dfrac{1}{b^2-bc+c^2}}< =\dfrac{1}{\sqrt{bc}};\sqrt{\dfrac{1}{c^2-ac+c^2}}< =\dfrac{1}{\sqrt{ac}}\)
=>P<=1/a+1/b+1/c=3
Dấu = xảy ra khi a=b=c=1
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\(ab+bc+ca=abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
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 :)
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\(\frac{ab}{c+1}=\frac{ab}{a+c+b+c}\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)
\(M\le\frac{1}{4}\left[\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{bc}{c+a}+\frac{ca}{a+b}+\frac{ca}{b+c}\right]\)
\(=\frac{1}{4}\left[\frac{b\left(a+c\right)}{a+c}+\frac{a\left(b+c\right)}{b+c}+\frac{c\left(a+b\right)}{a+b}\right]=\frac{1}{4}\left(a+b+c\right)=\frac{1}{4}\)
chịu thôi chị ơi!
Ai trả lời câu này được bái luôn thành sư phụ!!!!!
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\(\sqrt{\frac{ab}{c+ab}}=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}=\sqrt{\frac{ab}{c^2+ab+bc+ca}}\)
\(=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\le\frac{1}{2}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)
Tương tự cho 2 BĐT còn lại r` cộng vào nhé
Cho a ; b ; c > 0 ; ab + bc + ac = 1
Tìm max : \(\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}-\dfrac{1}{c^2+1}\)
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ĐK : a;b;c > 0
Ta có : \(ab+bc+ac=1\) \(\Leftrightarrow c\left(a+b\right)=1-ab\Leftrightarrow c=\dfrac{1-ab}{a+b}\)
Khi đó : \(c^2+1=\left(\dfrac{1-ab}{a+b}\right)^2+1\) \(=\dfrac{\left(ab\right)^2+1+a^2+b^2}{\left(a+b\right)^2}=\dfrac{\left(a^2+1\right)\left(b^2+1\right)}{\left(a+b\right)^2}\)
\(\Rightarrow\dfrac{1}{c^2+1}=\dfrac{\left(a+b\right)^2}{\left(a^2+1\right)\left(b^2+1\right)}\)
Ta có : \(\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}=\dfrac{ab^2+a^2b+a+b}{\left(a^2+1\right)\left(b^2+1\right)}=\dfrac{\left(ab+1\right)\left(a+b\right)}{\left(a^2+1\right)\left(b^2+1\right)}\)
Suy ra : \(A=\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}-\dfrac{1}{c^2+1}=\dfrac{\left(a+b\right)\left(ab+1-a-b\right)}{\left(a^2+1\right)\left(b^2+1\right)}=\dfrac{\left(a+b\right)\left(1-a\right)\left(1-b\right)}{\left(a^2+1\right)\left(b^2+1\right)}\)
AD BĐT Cauchy ta được : \(\left(a+b\right)\left[\left(1-a\right)\left(1-b\right)\right]\le\dfrac{\left[a+b+\left(1-a\right)\left(1-b\right)\right]^2}{4}=\dfrac{\left(1+ab\right)^2}{4}\)
\(\left(a^2+1\right)\left(b^2+1\right)\ge\left(ab+1\right)^2\) ( theo BCS )
Suy ra : \(A\le\dfrac{1}{4}\)