sai đề

Áp dụng BĐT Svacxo ta có :

\(\frac{1}{a^3\left(7b+3c\right)}+\frac{1}{b^3\left(7c+3a\right)}+\frac{1}{c^3\left(7a+3b\right)}=\frac{\frac{1}{a^2}}{7ab+7ac}+\frac{\frac{1}{b^2}}{7bc+3ab}+\frac{\frac{1}{c^2}}{7ac+3bc}\)

\(\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{10\left(ab+bc+ca\right)}=\frac{1}{10}.\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{ab+bc+ca}=\frac{1}{10}.\left(ab+bc+ca\right)\)

\(=\frac{1}{10}.\frac{ab+bc+ca}{abc}=\frac{1}{10}.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(đpcm\right)\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)

Áp dụng BĐT AM-GM ta có \(\frac{1^2}{a\left(a+b\right)}+\frac{1^2}{b\left(b+c\right)}+\frac{1^2}{c\left(c+a\right)}\ge\)

\(\ge\frac{\left(1+1+1\right)^2}{a\left(a+b\right)+b\left(b+c\right)+c\left(c+a\right)}=\frac{9}{a\left(a+b\right)+b\left(b+c\right)+c\left(c+a\right)}\ge\)

\(\ge\frac{9}{3.\sqrt[3]{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

HÌNH NHƯ NGƯỢC DẦU RỒI THÌ PHẢI

Bạn tham khảo tại đây:

Câu hỏi của Trần Hữu Ngọc Minh - Toán lớp 9 - Học toán với OnlineMath

Áp dụng BĐT Cosi ta được:

\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{8}+\frac{1+c}{8}\ge3\sqrt{\frac{a^3\left(1+b\right)\left(1+c\right)}{\left(1+b\right)\left(1+c\right)64}}=\frac{3a}{4}̸\)

Tương tự \(\hept{\begin{cases}\frac{b^3}{\left(1+a\right)\left(1+c\right)}+\frac{1+a}{8}+\frac{1+c}{8}\ge\frac{3b}{4}\\\frac{c^3}{\left(1+a\right)\left(1+b\right)}+\frac{1+a}{8}+\frac{1+b}{8}\ge\frac{3c}{4}\end{cases}}\)

Cộng theo từng vế BĐT trên ta có:

\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+a\right)\left(1+c\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}+\frac{3}{4}\ge\frac{a+b+c}{2}\)

Vì \(a+b+c\ge3\sqrt[3]{abc}=3\)do đó:

\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+a\right)\left(1+c\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}+\frac{3}{4}\ge\frac{3}{2}\)

\(\Leftrightarrow\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+a\right)\left(1+c\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{3}{4}\left(đpcm\right)\)

Đẳng thức xảy ra <=> a=b=c

bđt \(\Leftrightarrow\)\(\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)\ge\left(\frac{10}{3}\right)^3abc\) (*) 

đặt \(\left(\sqrt{ab};\sqrt{bc};\sqrt{ca}\right)=\left(x;y;z\right)\)\(\Rightarrow\)\(xyz\le\frac{1}{27}\)

(*) \(\Leftrightarrow\)\(\left(x^2+1\right)\left(y^2+1\right)\left(z^2+1\right)\ge\left(\frac{10}{3}\right)^3xyz\)

\(VT\ge\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)\)

Có \(xy+1\ge10\sqrt[10]{\frac{xy}{9^9}}\)

Tương tự với \(yz+1\)\(;\)\(zx+1\)\(\Rightarrow\)\(VT\ge10^3\sqrt[10]{\frac{\left(xyz\right)^2}{9^{27}}}\)

Ta cần CM \(10^3\sqrt[10]{\frac{\left(xyz\right)^2}{9^{27}}}\ge\frac{10^3}{3^3}xyz\) đúng với \(xyz\le\frac{1}{27}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)

Đặt \(P=\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)\)

Vì a+b+c=1 nên 

\(P=\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)=abc+\frac{1}{abc}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\)

Từ BĐt Cosi cho 3 số dương ta có:

\(\frac{1}{3}=\frac{a+b+c}{3}\ge\sqrt[3]{abc}\Rightarrow abc\le\frac{1}{27}\)

đặt x=abc thì \(0< x\le\frac{1}{27}\)

do đó: \(x+\frac{1}{x}-27-\frac{1}{27}=\frac{\left(27-x\right)\left(1-27x\right)}{27x}\ge0\)

=> \(x+\frac{1}{x}=abc+\frac{1}{abc}\ge27+\frac{1}{27}=\frac{730}{27}\)

Mặt khác: \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)

Nên  \(P\ge\frac{730}{27}+10=\frac{1000}{27}=\left(\frac{10}{3}\right)^3\)

Dấu "=" xảy ra khi a=b=c\(=\frac{1}{3}\)

Ta có:

\(\frac{a}{\left(a+1\right)\left(b+1\right)}+\frac{a\left(a+1\right)}{8}+\frac{a\left(b+1\right)}{8}\ge3\sqrt[3]{\frac{a^3\left(a+1\right)\left(b+1\right)}{64\left(a+1\right)\left(b+1\right)}}=\frac{3a}{4}\)

\(\Rightarrow LHS+\frac{a^2+b^2+c^2+ab+bc+ca+2\left(a+b+c\right)}{8}\ge\frac{3}{4}\left(a+b+c\right)\)

\(\Rightarrow LHS\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{4}\left(a+b+c\right)-\frac{a^2+b^2+c^2+ab+bc+ca}{8}\)

\(\ge\frac{a+b+c}{2}-\frac{a^2+b^2+c^2}{4}\)

Có ý tưởng đến đây thôi nhưng lại bị ngược dấu rồi :(

BĐT <=> \(\frac{a\left(c+1\right)+b\left(a+1\right)+c\left(b+1\right)}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\frac{3}{4}\)

<=> \(\frac{ab+bc+ac+a+b+c}{abc+1+ab+bc+ac+a+c+b}\ge\frac{3}{4}\)

<=> \(4\left(ab+bc+ac+a+b+c\right)\ge3\left(ab+bc+ac+a+b+c+2\right)\)

<=> \(ab+bc+ac+a+b+c\ge6\)(1)

(1) luôn đúng do \(ab+bc+ac\ge3\sqrt[3]{a^2b^2c^2}=3;a+b+c\ge3\sqrt[3]{abc}=3\)

=> BĐT được CM

Dấu bằng xảy ra khi \(a=b=c=1\)

Biến đổi tương đương ta có : 

\(\frac{a}{\left(a+1\right).\left(b+1\right)}+\frac{b}{\left(b+1\right).\left(c+1\right)}+\frac{c}{\left(c+1\right).\left(a+1\right)}\ge\frac{3}{4}\)

\(\Leftrightarrow4.a.\left(c+1\right)+4.b.\left(a+1\right)+4.c.\left(b+1\right)\ge3.\left(a+1\right).\left(b+1\right).\left(c+1\right)\)

\(\Leftrightarrow4.\left(a+b+c\right)+4.\left(ab+bc+ac\right)\ge3.a.b.c+3.\left(a+b+c\right)+3.\left(ab+bc+ca\right)+3\)

\(\Leftrightarrow a+b+c+ab+bc+ca\ge6\)

Sử dụng thêm bất đẳng thức Cauchy 3 số ta có : 

a+b+c \(\ge\)3.\(\sqrt[3]{abc}\)và ab + bc + ca \(\ge3.\sqrt[3]{a^2b^2c^2}=3\)

Vậy bất đẳng thức đã được chứng minh . Dấu bằng xảy ra khi và chỉ khi a= b= c =1

Mình áp dụng BĐT AM-GM  đến dòng 

\(\Leftrightarrow ab+bc+ca+a+b\ge6\left(1\right)\)

Áp dụng BĐT AM-GM cho 3 số dương ta được

\(ab+bc+ca\ge3\sqrt[2]{\left(abc\right)^2}=3;a+b+c\ge3\sqrt[2]{abc}=3\)

Cộng từng vế  BĐT ta được (1). Do vậy BĐT ban đầu được chứng minh

Dấu "=" xảy ra <=> a=b=c=1

Biến đối tương đương ta có:

\(\frac{a}{\left(a+1\right)\left(b+1\right)}+\frac{b}{\left(b+1\right)\left(c+1\right)}+\frac{c}{\left(c+1\right)\left(a+1\right)}\ge\frac{3}{4}\)

\(\Leftrightarrow4a\left(c+1\right)+4b\left(a+1\right)+4c\left(b+1\right)\ge3\left(a+1\right)\left(b+1\right)\left(c+1\right)\)

\(\Leftrightarrow4\left(a+b+c\right)+4\left(ab+bc+ca\right)\ge3abc+3\left(a+b+c\right)+3\left(ab+bc+ca\right)+3\)

\(\Leftrightarrow a+b+c+ab+bc+ca\ge6\)

Sử dụng thêm BĐT Cauchy 3 số ta có:

\(\hept{\begin{cases}a+b+c\ge3\sqrt[3]{abc}=3\\ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}=3\end{cases}}\)

Vậy BĐT đã được chứng minh. Dấu "=" <=> a=b=c=1