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BĐT \(\Leftrightarrow\left[\left(a+b\right)+\left(a+c\right)\right]\left[\left(b+c\right)+\left(a+b\right)\right]\left[\left(c+a\right)+\left(b+c\right)\right]\ge8\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Đây là BĐT quy thuộc! \(\left(a+b\right)+\left(a+c\right)\ge2\sqrt{\left(a+b\right)\left(a+c\right)}\) rồi tương tự các kiểu.
Nhân theo vế thu được đpcm
Sử dụng BĐT: \(\left(x+y+z\right)^3\ge27xyz\Rightarrow\left(\frac{x+y+z}{3}\right)^3\ge xyz\)
\(\Rightarrow\left(\frac{1+a+1+b+1+c}{3}\right)^3\ge\left(1+a\right)\left(1+b\right)\left(1+c\right)\)
Ta có: \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge3\sqrt[3]{\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
\(\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}\ge3\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
Cộng vế với vế:
\(1\ge\frac{1+\sqrt[3]{abc}}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\Rightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)
Dấu "=" 3 BĐT trên xảy ra khi \(a=b=c\)
Lại có:
\(1+\sqrt[3]{abc}\ge2\sqrt{\sqrt[3]{abc}}\Rightarrow\left(1+\sqrt[3]{abc}\right)^3\ge\left(2\sqrt{\sqrt[3]{abc}}\right)^3=8\sqrt{abc}\)Dấu "=" xảy ra khi \(a=b=c=1\)
\(\frac{1}{a}-1=\frac{a+b+c}{a}-\frac{a}{a}=\frac{b+c}{a}\)
Tương tự : \(\frac{1}{b}-1=\frac{c+a}{b};\frac{1}{c}-1=\frac{a+b}{c}\)
Nhân theo vế ta đc :
\(VT=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Áp dụng bđt Cauchy :
\(VT\ge\frac{8abc}{abc}=8\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
\(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)
\(=2\left(1+abc\right)+\sqrt{\left[\left(a+1\right)^2+\left(1-a\right)^2\right]\left[\left(b+c\right)^2+\left(bc-1\right)^2\right]}\)
\(\ge2\left(1+abc\right)+\left(a+1\right)\left(b+c\right)+\left(1-a\right)\left(bc-1\right)\)
\(=\left(1+a\right)\left(1+b\right)\left(1+c\right)\)
\(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}.\)
\(=2\left(1+abc\right)+\sqrt{\left[\left(a+1\right)^2+\left(1-a\right)^2\right]\left[\left(b+c\right)^2+\left(bc-1\right)^2\right]}\)
\(\ge2\left(1+abc\right)+\left(a+1\right)\left(b+c\right)+\left(1-a\right)\left(bc-1\right)\)
\(=\left(1+a\right)\left(1+b\right)\left(1+c\right)\)
Từ \(a+b+c=1\) thế vào biểu thức sau
\(\left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right)\left(\frac{1}{c}-1\right)=\left(\frac{a+b+c}{a}-\frac{a}{a}\right)\left(\frac{a+b+c}{b}-\frac{b}{b}\right)\left(\frac{a+b+c}{c}-\frac{c}{c}\right)\)
\(=\frac{b+c}{a}.\frac{a+c}{b}.\frac{a+b}{c}=\frac{\left(a+b\right)\left(b+c\right)\left(a+c\right)}{abc}\)(1)
Với a,b,c>0 , Áp dụng bất đẳng thức AM-GM (cauchy) cho hai số không âm ta có:
\(a+b\ge2\sqrt{ab};b+c\ge2\sqrt{bc};a+c\ge2\sqrt{ac}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)\ge2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ac}=8abc\)(2)
Từ (1) và (2) suy ra \(\left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right)\left(\frac{1}{c}-1\right)\ge\frac{8abc}{abc}=8\)
Dấu '=' xảy ra khi \(\hept{\begin{cases}a=b=c\\a+b+c=1\end{cases}\Leftrightarrow}a=b=c=\frac{1}{3}\)
Giải:
\(\dfrac{a}{\left(a+1\right)\left(b+1\right)}+\dfrac{b}{\left(b+1\right)\left(c+1\right)}+\dfrac{c}{\left(c+1\right)\left(a+1\right)}\ge\dfrac{3}{4}\)(*)
\(\Leftrightarrow\) \(\dfrac{a\left(c+1\right)+b\left(a+1\right)+c\left(b+1\right)}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{3}{4}\)
\(\Leftrightarrow\) \(\dfrac{ac+a+ab+b+bc+c}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\) \(\ge\) \(\dfrac{3}{4}\)
Do a+1 ; b+1; c+1 >0
\(\Rightarrow\) 4ac+4a+4ab+4b+4bc+4c \(\ge\) 3abc+3ac+3bc+3ab+3a+3b+3c+3
\(\Leftrightarrow\) ac+ab+bc+a+b+c -6 \(\ge\) 0
Áp dụng BĐT Cô-si cho 3 số
Ta có: a+b+c \(\ge\) \(3\sqrt[3]{abc}=3\)
ab+bc+ca \(\ge\) \(3\sqrt[3]{\left(abc\right)^2}\) = 3
\(\Rightarrow\)ac+ab+bc+a+b+c -6 \(\ge\) 0 ( luôn đúng)
\(\Rightarrow\) (*) được chứng minh
Dấu "=" xảy ra \(\Leftrightarrow\) a=b=c=1
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 \(a=b=c=1\)