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Từ giả thiết suy ra \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\) (*) (Vì a,b,c > 0)
Áp dụng BĐT Cauchy ta có:
\(\frac{1}{\sqrt{a^3+b}}\le\frac{1}{\sqrt{2}.\sqrt[4]{a^3b}}=\frac{1}{\sqrt{2}}.\sqrt[4]{\frac{1}{a}.\frac{1}{a}.\frac{1}{a}.\frac{1}{b}}\le\frac{1}{4\sqrt{2}}\left(\frac{3}{a}+\frac{1}{b}\right)\)
Đánh giá tương tự: \(\frac{1}{\sqrt{b^3+c}}\le\frac{1}{4\sqrt{2}}\left(\frac{3}{b}+\frac{1}{c}\right);\frac{1}{\sqrt{c^3+a}}\le\frac{1}{4\sqrt{2}}\left(\frac{3}{c}+\frac{1}{a}\right)\)
Từ đó, kết hợp với (*) suy ra:
\(\frac{1}{\sqrt{a^3+b}}+\frac{1}{\sqrt{b^3+c}}+\frac{1}{\sqrt{c^3+a}}\le\frac{1}{4\sqrt{2}}.4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{3\sqrt{2}}{2}\)(đpcm)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=1.\)
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a) Ta có BĐT:
\(a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\ge\left(a+b\right)ab\)
\(\Rightarrow a^3+b^3+abc\ge ab\left(a+b+c\right)\)
\(\Rightarrow\frac{1}{a^3+b^3+abc}\le\frac{1}{ab\left(a+b+c\right)}\)
Tương tự cho 2 bất đẳng thức còn lại rồi cộng theo vế:
\(VT\le\frac{1}{ab\left(a+b+c\right)}+\frac{1}{bc\left(a+b+c\right)}+\frac{1}{ca\left(a+b+c\right)}\)
\(=\frac{a+b+c}{abc\left(a+b+c\right)}=\frac{1}{abc}=VP\)
Khi \(a=b=c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(\frac{bc}{\sqrt{a+bc}}=\frac{bc}{\sqrt{a\left(a+b+c\right)+bc}}=\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(=\sqrt{\frac{bc}{a+b}}.\sqrt{\frac{bc}{a+c}}\le\frac{1}{2}\left(\frac{bc}{a+b}+\frac{bc}{a+c}\right)\)
Tương tự với hai BĐT còn lại và cộng theo vế rồi rút gọn ta được \(VT\le\frac{a+b+c}{2}=\frac{1}{2}\left(đpcm\right)\)
Đẳng thức xảy ra khi a= b=c=1/3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{a+bc}=\sqrt{a\left(a+b+c\right)+bc}=\sqrt{a^2+ab+ac+bc}\)
\(=\sqrt{a\left(a+b\right)+c\left(a+b\right)}=\sqrt{\left(a+b\right)\left(a+c\right)}\)
\(\Rightarrow\frac{bc}{\sqrt{a+bc}}=\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}=\sqrt{\frac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\)
Áp dụng bđt Cô-si :
\(\sqrt{\frac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\le\frac{\frac{bc}{a+b}+\frac{bc}{a+c}}{2}\)
Chứng minh tương tự với các phân thức còn lại, cộng theo vế ta có :
\(VT\le\frac{\left(\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ac}{c+b}+\frac{ac}{a+b}+\frac{ab}{a+c}+\frac{ab}{b+c}\right)}{2}\)
\(=\frac{\frac{c\left(a+b\right)}{a+b}+\frac{b\left(a+c\right)}{a+c}+\frac{a\left(b+c\right)}{b+c}}{2}=\frac{a+b+c}{2}=\frac{1}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{\frac{ab}{c+ab}}=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}=\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+c}+\frac{b}{b+c}\right)\)
Tương tự: \(\sqrt{\frac{bc}{a+bc}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\) ; \(\sqrt{\frac{ca}{b+ca}}\le\frac{1}{2}\left(\frac{c}{b+c}+\frac{a}{a+b}\right)\)
Cộng vế với vế: \(VT\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{b+c}+\frac{a}{a+c}+\frac{c}{a+c}\right)=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bất đẳng thức bu nhi a ta có \(\left(x^2+y^2+z^2\right)3\ge\left(x+y+z\right)^2\)
Áp dụng ta có
\(Q^2\le3\left(\frac{a}{1+a+ab}+\frac{b}{1+b+bc}+\frac{c}{1+c+ca}\right)\)
đặt \(M=\frac{a}{1+a+ab}+\frac{b}{1+b+bc}+\frac{c}{1+c+ca}=\frac{a}{1+a+ab}+\frac{ab}{a+ab+abc}+\frac{abc}{ab+abc+â^2bc}\)
\(=\frac{1}{a+ab+1}+\frac{a}{a+ab+1}+\frac{ab}{1+ab+1}=1\)
=> \(Q^2\le3\Rightarrow Q\le\sqrt{3}\)
mặt khác Áp dụng cô si ta có
\(a+b+c\ge3\sqrt[3]{abc}=3\Rightarrow\sqrt{a+b+c}\ge\sqrt{3}\Rightarrow\sqrt{a+b+c}\ge Q\) (ĐPCM)
ta có:
\(\frac{a}{1+a+ab}+\frac{b}{1+b+bc}+\frac{c}{1+c+ca}=\frac{a}{abc+a+ab}+\frac{b}{1+b+bc}+\frac{bc}{b+bc+abc}\)
\(=\frac{1}{1+b+bc}+\frac{b}{1+b+bc}+\frac{bc}{1+b+bc}=1\)
ta có:
\(Q^2\le3\left(\frac{a}{1+a+ab}+\frac{b}{1+b+bc}+\frac{c}{1+c+ca}\right)=3\)
\(\Rightarrow Q\le\sqrt{3}=\sqrt{3\sqrt[3]{abc}}\le\sqrt{a+b+c}\left(Q.E.D\right)\)
dấu = xảy ra khi a=b=c=1
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(a^2=x;b^2=y;c^2=z\)
Ta có:
\(VT=\sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}\)
Mặt khác:
\(\sqrt{\frac{x}{x+y}}=\sqrt{\frac{x}{\left(x+y\right)\left(x+z\right)}\cdot\sqrt{x+z}}\)
Áp dụng Bđt Cauchy-Schwarz ta có:
\(VT^2\le2\left[\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(y+z\right)\left(y+x\right)}+\frac{z}{\left(z+x\right)\left(z+y\right)}\right]\left(x+y+z\right)\)
\(\Leftrightarrow VT^2\le\frac{4\left(x+y+z\right)\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Vì \(VP^2=\frac{9}{2}\) nên cần chứng minh \(VT^2\le\frac{9}{2}\)
\(\Leftrightarrow9\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8\left(x+y+z\right)\left(xy+yz+zx\right)\)
bn tự lm tiếp
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(\sqrt{\frac{ab}{c+ab}}=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}\)\(=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\)\(=\sqrt{\frac{a}{c+a}}.\sqrt{\frac{b}{c+b}}\)\(\le\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)
Tương tự, ta có
\(\sqrt{\frac{bc}{a+bc}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\)
\(\sqrt{\frac{ca}{b+ca}\le\frac{1}{2}\left(\frac{c}{c+b}+\frac{a}{b+a}\right)}\)
Cộng vế theo vế của 3 bđt ta được đpcm