\(abc=1\ge a^2b^2c^2=1\)
\(\Rightarrow\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}=\frac{b^2c^2}{a\left(b+c\right)}+\frac{a^2c^2}{b\left(c+a\right)}+\frac{a^2b^2}{c\left(a+b\right)}\)
Theo Cauchy-Schwarz ta được:
\(VP\ge\frac{\left(bc+ab+ac\right)^2}{2\left(ab+ac+bc\right)}=\frac{bc+ab+ac}{2}\ge\frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi a = b = c = 1
\(\)
Bài làm:
Ta có: \(\frac{1}{a^3\left(b+c\right)}=\frac{abc}{a^3\left(b+c\right)}=\frac{bc}{a^2b+a^2c}\)
\(=\frac{b^2c^2}{a^2b^2c+a^2bc^2}=\frac{b^2c^2}{ab+ac}\)
Tương tự: \(\frac{1}{b^3\left(c+a\right)}=\frac{c^2a^2}{ba+bc}\) ; \(\frac{1}{c^3\left(a+b\right)}=\frac{a^2b^2}{ca+cb}\)
=> \(Vt=\frac{a^2b^2}{ca+bc}+\frac{b^2c^2}{ab+ca}+\frac{c^2a^2}{ab+bc}\ge\frac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi: \(a=b=c=1\)
Cách khác:
\(Vt=\frac{\frac{1}{a^2}}{a\left(b+c\right)}+\frac{\frac{1}{b^2}}{b\left(c+a\right)}+\frac{\frac{1}{c^2}}{c\left(a+b\right)}\)
\(\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(\frac{ab+bc+ca}{abc}\right)^2}{2\left(ab+bc+ca\right)}\)
\(=\frac{ab+bc+ca}{2abc}\ge\frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi: a = b = c = 1
Cách khác :">
\(\left(a,b,c\right)->\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\Rightarrow xyz=1\)
\(\Rightarrow VT=\frac{x^3yz}{y+z}+\frac{y^3xz}{x+z}+\frac{z^3xy}{x+y}\)
\(=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}\)
\(=\frac{x+y+z}{2}\ge\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z=1\Leftrightarrow a=b=c=1\)
hơi dài do mình chứng minh bđt phụ nhé ( cách giải không dùng bđt svacxo nhưng có dùng AM-GM )
Trước tiên ta chứng minh BĐT : Với \(\forall a,b,c\inℝ\)và \(x,y,z>0\)ta có :
\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)(*)
Dấu = xảy ra \(< =>\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
Thật vậy , với \(a,b\inℝ\)và \(x,y>0\)ta có :
\(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\)(**)
\(< =>\left(a^2y+b^2x\right)\left(x+y\right)\ge xy\left(a+b\right)^2\)
\(< =>\left(bx-ay\right)^2\ge0\)*đúng*
Dấu = xảy ra \(< =>\frac{a}{x}=\frac{b}{y}\)
Áp dụng liên tiếp bất đẳng thức (**) ta có :
\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b\right)}{x+y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
Dấu = xảy ra \(< =>\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
Quay trở lại bài toán ta có : \(LHS=\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ba}+\frac{\frac{1}{c^2}}{ca+cb}\)
Áp dụng bất đẳng thức (*) ta có :
\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{ba+bc}+\frac{\frac{1}{c^2}}{ca+cb}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}=\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{2}\)
Hay \(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{ba+bc}+\frac{\frac{1}{c^2}}{ca+cb}\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Bài toán sẽ hoàn tất nếu ta chỉ ra được \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\)
Thật vậy \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}=3\left(QED\right)\)
Vậy ta được \(LHS\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\frac{1}{2}.3=\frac{3}{2}\)
Dấu = xảy ra khi và chỉ khi \(a=b=c=1\)
