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Bài 1 :
a) Ta có : \(\left(1-a\right)\left(1-b\right)\left(1-c\right)=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Áp dụng bđt Cauchy : \(a+b\ge2\sqrt{ab}\) , \(b+c\ge2\sqrt{bc}\) , \(c+a\ge2\sqrt{ca}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\) hay \(\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge8abc\)
Áp dụng bđt Cô-si :
\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab\cdot bc}{ca}}=2b\)
Tương tự : \(\frac{bc}{a}+\frac{ca}{b}\ge2c;\frac{ab}{c}+\frac{ca}{b}\ge2a\)
Cộng theo vế 3 bđt :
\(2\cdot\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
Đề đúng : Cho a,b,c > 0 và \(a+b+c\le1\)
CMR : \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge9\)
Đặt \(x=a^2+2bc,y=b^2+2ac,z=c^2+2ab\)
Áp dụng bđt Bunhiacopxki , ta có: \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(x+y+z\right)\ge\left(\sqrt{\frac{1}{x}.x}+\sqrt{\frac{1}{y}.y}+\sqrt{\frac{1}{z}.z}\right)^2=9\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\) hay \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge\frac{9}{\left(a+b+c\right)^2}\ge9\)
Ta thấy: \(\left(a^2+2bc\right)+\left(b^2+2ac\right)+\left(c^2+2ab\right)=\left(a+b+c\right)^2\le1\)
Sử dụng Cosi 3 số ta suy ra
\(VT\ge\left[\left(a^2+2bc\right)+\left(b^2+2ac\right)+\left(c^2+2ab\right)\right]\left(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\right)\)
\(\ge3\sqrt[3]{\left(a^2+2bc\right)\left(b^2+2ac\right)\left(c^2+2ab\right)}\cdot3\sqrt[3]{\frac{1}{a^2+2bc}\cdot\frac{1}{b^2+2ac}\cdot\frac{1}{c^2+2ab}}=9\) (Đpcm)
Đẳng thức xảy ra khi\(\begin{cases}a+b+c=1\\a^2+2bc=b^2+2ac=c^2+2ab\end{cases}\)\(\Leftrightarrow a=b=c=\frac{1}{3}\)
\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}.\frac{bc}{a}}=2b\)
tương tự cộng theo vế rút gọn ta có đpcm
\(\left(1+\frac{b^2+c^2-a^2}{2bc}\right).\frac{1+\frac{a}{b+c}}{1-\frac{a}{b+c}}.\frac{b^2+c^2-\left(b-c\right)^2}{a+b+c}\)
= \(\left(1+\frac{\left(b+c\right)^2-2bc-a^2}{2bc}\right).\frac{\frac{a+b+c}{b+c}}{\frac{b+c-a}{b+c}}.\frac{\left(b+c\right)^2-2bc-\left(b-c\right)^2}{a+b+c}\)
= \(\left(1+\frac{\left(b+c-a\right)\left(b+c+a\right)-2bc}{2bc}\right).\frac{a+b+c}{b+c-a}.\frac{\left(b+c-b+c\right)\left(b+c+b-c\right)-2bc}{a+b+c}\)
= \(\left(1+\frac{\left(b+c-a\right)\left(b+c+a\right)}{2bc}-1\right).\frac{a+b+c}{b+c-a}.\frac{4bc-2bc}{a+b+c}\)
= \(\frac{\left(b+c-a\right)\left(b+c+a\right)}{2bc}.\frac{2bc}{b+c-a}\)
= \(\frac{\left(b+c-a\right)\left(b+c+a\right)}{b+c-a}\)
= \(b+c+a\)