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Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)(ab+bc+ac)\geq (a+b+c)^2\)
\(\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\geq \left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right)^2\)
Nhân theo vế 2 BĐT trên:
\(\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2(ab+bc+ac).\frac{a+b+c}{abc}\geq [(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})]^2\)
\(\Leftrightarrow \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq [(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})]^2\)
\(\Leftrightarrow \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\geq (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
TL :
Bất đẳng thức sai, chẳng hạn với \(a=b=10^{-4};c=0,5-a-b.\).
HT
Thưa anh, nếu \(a=b=10^{-4}\) và \(c=0,5-a-b=0,5-2.10^{-4}\),em bấm máy thì ngay cả khi chỉ có một cái
\(\frac{1}{ab\left(a+b\right)}\)nó đã bằng \(5.10^{11}\)lớn hơn rất nhiều so với \(\frac{87}{2}\), BĐT vẫn đúng chứ ạ?
ta có: \(\frac{a}{\left(a+1\right)\left(b+1\right)}+\frac{b}{\left(b+1\right)\left(c+1\right)}+\frac{c}{\left(c+1\right)\left(a+1\right)}.\)
\(\ge3\sqrt[3]{\frac{a.b.c}{\left(a+1\right)^2.\left(b+1\right)^2.\left(c+1\right)^2}}=\frac{3}{\sqrt[3]{\left(a+1\right)^2.\left(b+1\right)^2.\left(c+1\right)^2}}\) (vì abc=1) (*)
Mặt khác: \(\left(a+1\right)^2.\left(b+1\right)^2.\left(c+1\right)^2\ge64abc=64=4^3\) (vì abc=1)
=> \(\sqrt[3]{\left(a+1\right)^2.\left(b+1\right)^2.\left(c+1\right)^2}\ge4\) (**)
Từ (*), (**)=> đpcm
Bạn dưới kia làm ngược dấu thì phải,mà bài này hình như là mũ 3
\(\frac{a^3}{\left(a+1\right)\left(b+1\right)}+\frac{a+1}{8}+\frac{b+1}{8}\ge3\sqrt[3]{\frac{a^3\left(a+1\right)\left(b+1\right)}{64\left(a+1\right)\left(b+1\right)}}=\frac{3a}{4}\)
Tương tự rồi cộng lại:
\(RHS+\frac{2\left(a+b+c\right)+6}{8}\ge\frac{3\left(a+b+c\right)}{4}\)
\(\Leftrightarrow RHS\ge\frac{3}{4}\) tại a=b=c=1
Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)
\(x+y+z\ge\frac{x^2+2xy}{2x+y}+\frac{y^2+2yz}{2y+z}+\frac{z^2+2zx}{2z+x}\)
\(\Leftrightarrow x+y+z\ge\frac{3xy}{2x+y}+\frac{3yz}{2y+z}+\frac{3zx}{2z+x}\)
\(\frac{3xy}{2x+y}\le\frac{3}{9}xy\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{3}\left(x+2y\right)\)
\(\Rightarrow\Sigma_{cyc}\frac{3xy}{2x+y}\le\frac{1}{3}\left[\left(x+2y\right)+\left(y+2z\right)+\left(z+2x\right)\right]=x+y+z\)
Dấu "=" xảy ra khi x=y=z
\(2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\)
Thay thế \(a+b+c=1\)
\(\Leftrightarrow2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge\frac{2a+b+c}{b+c}+\frac{a+2b+c}{a+c}+\frac{a+b+2c}{a+b}\)
\(\Leftrightarrow2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge\frac{2a}{b+c}+\frac{2b}{a+c}+\frac{2c}{a+b}+3\)
\(\Leftrightarrow\frac{2b}{a}+\frac{2c}{b}+\frac{2a}{c}\ge\frac{2a}{b+c}+\frac{2b}{a+c}+\frac{2c}{a+b}+3\)
\(\Leftrightarrow\left(\frac{2b}{a}-\frac{2b}{a+c}\right)+\left(\frac{2c}{b}-\frac{2c}{a+b}\right)+\left(\frac{2a}{c}-\frac{2a}{b+c}\right)\ge3\)
\(\Leftrightarrow\frac{2bc}{a\left(a+c\right)}+\frac{2ca}{b\left(a+b\right)}+\frac{2ab}{c\left(b+c\right)}\ge3\)
\(\Leftrightarrow\frac{bc}{a\left(a+c\right)}+\frac{ca}{b\left(a+b\right)}+\frac{ab}{c\left(b+c\right)}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{\left(bc\right)^2}{abc\left(a+c\right)}+\frac{\left(ca\right)^2}{abc\left(a+b\right)}+\frac{\left(ab\right)^2}{abc\left(b+c\right)}\ge\frac{3}{2}\)
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow\frac{\left(bc\right)^2}{abc\left(a+c\right)}+\frac{\left(ca\right)^2}{abc\left(a+b\right)}+\frac{\left(ab\right)^2}{abc\left(b+c\right)}\)
\(\ge\frac{\left(ab+bc+ca\right)^2}{abc\left(a+b+c+a+b+c\right)}=\frac{\left(ab+bc+ca\right)^2}{2abc}\)
Chứng minh rằng : \(\frac{\left(ab+bc+ca\right)^2}{2abc}\ge\frac{3}{2}\)
\(\Leftrightarrow2\left(ab+bc+ca\right)^2\ge6abc\)
\(\Leftrightarrow\left(ab+bc+ca\right)^2\ge3abc\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc\ge3abc\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\ge3abc\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\ge3abc\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\hept{\begin{cases}a^2b^2+b^2c^2\ge2\sqrt{a^2b^4c^2}=2ab^2c\\b^2c^2+c^2a^2\ge2\sqrt{a^2b^2c^4}=2abc^2\\a^2b^2+c^2a^2\ge2\sqrt{a^2b^2c^2}=2a^2bc\end{cases}}\)
\(\Leftrightarrow2\left(a^2b^2+b^2c^2+c^2a^2\right)\ge2abc\left(a+b+c\right)\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\left(đpcm\right)\)
Vì \(\frac{\left(ab+bc+ca\right)^2}{2abc}\ge\frac{3}{2}\)
Vậy \(\frac{\left(bc\right)^2}{abc\left(a+c\right)}+\frac{\left(ca\right)^2}{abc\left(a+b\right)}+\frac{\left(ab\right)^2}{abc\left(b+c\right)}\ge\frac{3}{2}\)
\(\Leftrightarrow2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\left(đpcm\right)\)
Chúc bạn học tốt !!!