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Các bạn giúp mình mấy câu BĐT Cauchy này với

1. cho a,b,c>0 và a+b+c=6 CMR $\frac{a}{\sqrt{b^3+1}}+\frac{b}{\sqrt{c^3+1}}+\frac{c}{\sqrt{a^3+1}}\ge2$

2.cho a,b,c>0 CMR \(\frac{ab}{\sqrt{c^2+3}}+\f...

Trần Phúc Khang · Accepted Answer

3.Áp dụng BĐT $\frac{1}{x+y+z}\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ta có

$\frac{ab}{a+3b+2c}=ab.\frac{1}{\left(a+c\right)+2b+\left(b+c\right)}\le\frac{1}{9}ab.\left(\frac{1}{a+c}+\frac{1}{2b}+\frac{1}{b+c}\right)$

TT $\frac{bc}{b+3c+2a}\le\frac{bc}{9}.\left(\frac{1}{b+a}+\frac{1}{2c}+\frac{1}{c+a}\right)$

$\frac{ca}{c+3a+2b}\le\frac{ac}{9}.\left(\frac{1}{a+b}+\frac{1}{2a}+\frac{1}{b+c}\right)$

=> $VT\le\frac{1}{18}\left(a+b+c\right)+\Sigma.\frac{1}{9}.\left(\frac{bc}{a+c}+\frac{ba}{a+c}\right)=\frac{1}{18}\left(a+b+c\right)+\frac{1}{9}\left(a+b+c\right)=\frac{1}{6}\left(a+b+c\right)$

Dấu bằng xảy ra khi a=b=cCâu trả lời: 3.Áp dụng BĐT $\frac{1}{x+y+z}\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ta có

$\frac{ab}{a+3b+2c}=ab.\frac{1}{\left(a+c\right)+2b+\left(b+c\right)}\le\frac{1}{9}ab.\left(\frac{1}{a+c}+\frac{1}{2b}+\frac{1}{b+c}\right)$

TT $\frac{bc}{b+3c+2a}\le\frac{bc}{9}.\left(\frac{1}{b+a}+\frac{1}{2c}+\frac{1}{c+a}\right)$

$\frac{ca}{c+3a+2b}\le\frac{ac}{9}.\left(\frac{1}{a+b}+\frac{1}{2a}+\frac{1}{b+c}\right)$

=> $VT\le\frac{1}{18}\left(a+b+c\right)+\Sigma.\frac{1}{9}.\left(\frac{bc}{a+c}+\frac{ba}{a+c}\right)=\frac{1}{18}\left(a+b+c\right)+\frac{1}{9}\left(a+b+c\right)=\frac{1}{6}\left(a+b+c\right)$

Dấu bằng xảy ra khi a=b=c

Trần Phúc Khang · Answer

2. Chuẩn hóa $a+b+c=3$

=> $ab+bc+ac\le3$

=> $c^2+3\ge\left(a+c\right)\left(b+c\right)$

=> $\frac{ab}{\sqrt{c^2+3}}\le\frac{ab}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{1}{2}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)$

=> $VT\le\Sigma\frac{1}{2}\left(\frac{ab}{a+c}+\frac{bc}{a+c}\right)=\frac{1}{2}\left(a+b+c\right)=\frac{3}{2}$(ĐPCM)

Dấu bằng xảy ra khi a=b=c=1Câu trả lời: 2. Chuẩn hóa $a+b+c=3$

=> $ab+bc+ac\le3$

=> $c^2+3\ge\left(a+c\right)\left(b+c\right)$

=> $\frac{ab}{\sqrt{c^2+3}}\le\frac{ab}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{1}{2}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)$

=> $VT\le\Sigma\frac{1}{2}\left(\frac{ab}{a+c}+\frac{bc}{a+c}\right)=\frac{1}{2}\left(a+b+c\right)=\frac{3}{2}$(ĐPCM)

Dấu bằng xảy ra khi a=b=c=1

Trần Phúc Khang · Answer

1. Ta có $\sqrt{b^3+1}=\sqrt{\left(b+1\right)\left(b^2-b+1\right)}\le\frac{1}{2}\left(b^2+2\right)$

=> $\frac{a}{\sqrt{b^3+1}}\ge\frac{2a}{2+b^2}=\frac{2a+ab^2-ab^2}{2+b^2}=a-\frac{2ab^2}{b^2+b^2+4}$

Lại có $b^2+b^2+4\ge3\sqrt[3]{b^4.4}$

=> $\frac{a}{\sqrt{b^3+1}}\ge a-\frac{2ab^2}{3\sqrt[3]{b^4.4}}=a-\frac{2}{3}.a.\sqrt[3]{\frac{b^2}{4}}$

Mà $\sqrt[3]{\frac{b^2}{4}.1}=\sqrt[3]{\frac{b}{2}.\frac{b}{2}.1}\le\frac{1}{3}\left(b+1\right)$

=>$\frac{a}{\sqrt[3]{b^3+1}}\ge a-\frac{2}{3}.a.\frac{1}{3}\left(b+1\right)=\frac{7a}{9}-\frac{2}{9}ab$

Khi đó

$VT\ge\frac{7}{9}\left(a+b+c\right)-\frac{2}{9}\left(ab+bc+ac\right)$

Mà $ab+bc+ac\le\frac{1}{3}\left(a+b+c\right)^2=12$

=> $VT\ge\frac{7}{9}.6-\frac{2}{9}.12=2$(ĐPCM)

Dấu bằng xảy ra khi a=b=c=2Câu trả lời: 1. Ta có $\sqrt{b^3+1}=\sqrt{\left(b+1\right)\left(b^2-b+1\right)}\le\frac{1}{2}\left(b^2+2\right)$