Áp dụng BĐT Cauchy Schwarz, BĐT AM - GM và BĐT \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\), ta có:
\(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(a+c\right)}+\dfrac{1}{c^3\left(a+b\right)}\)
\(\ge\dfrac{\left(1+1+1\right)^2}{a^3\left(b+c\right)+b^3\left(a+c\right)+c^3\left(a+b\right)}\)
\(\ge\dfrac{3^2}{3\sqrt[3]{a^3b^3c^3\left(b+c\right)\left(a+c\right)\left(a+b\right)}}\)
\(\ge\dfrac{3^2}{3\sqrt[3]{8abc}}=\dfrac{3}{2}\left(abc=1\right)\)
Dấu "=" xảy ra khi a = b = c = 1
Với mọi \(a,b,c\in R\) và \(x,y,z>0\) . ta có:
\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\ge\dfrac{\left(a+b+c\right)^2}{x+y+z}\) (1)
Dấu "=" xảy ra \(\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
Thật vậy, với \(a,b\in R\) và \(x,y>0\) ta có
\(\dfrac{a^2}{x}+\dfrac{b^2}{y}\ge\dfrac{\left(a+b\right)^2}{x+y}\) (2)
\(\Leftrightarrow\left(a^2y+b^2x\right)\left(x+y\right)\ge xy\left(a+b\right)^2\)
\(\Leftrightarrow\left(bx-ay\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra \(\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}\)
Áp dụng BĐT (2) ta có
\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\ge\dfrac{\left(a+b\right)^2}{x+y}+\dfrac{c^2}{z}\ge\dfrac{\left(a+b+c\right)^2}{x+y+z}\)Dấu "=" xảy ra\(\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
Ta có:
\(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(a+c\right)}+\dfrac{1}{c^3\left(a+b\right)}=\dfrac{\dfrac{1}{a^2}}{ab+ac}+\dfrac{\dfrac{1}{b^2}}{bc+ab}+\dfrac{\dfrac{1}{c^2}}{ca+cb}\)Áp dụng BĐT (1) ta có:
\(\dfrac{\dfrac{1}{a^2}}{ab+ac}+\dfrac{\dfrac{1}{b^2}}{ab+bc}+\dfrac{\dfrac{1}{c^2}}{ac+bc}\ge\dfrac{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}{2\left(ab+bc+ac\right)}=\dfrac{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}{2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)}\)(vì abc=1)
Hay \(\dfrac{\dfrac{1}{a^2}}{ab+bc}+\dfrac{\dfrac{1}{b^2}}{ab+bc}+\dfrac{\dfrac{1}{c^2}}{ac+bc}\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)Mà \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\) nên \(\dfrac{\dfrac{1}{a^2}}{ab+ac}+\dfrac{\dfrac{1}{b^2}}{ab+bc}+\dfrac{\dfrac{1}{c^2}}{ac+bc}\ge\dfrac{3}{2}\Rightarrowđpcm\)
đổi biến x=1/a,y=1/b,z=1/c=> xyz=1 bất đẳng thức cần chứng minh tương đương \(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\) dấu''='' xảy ra =>x=y=z=1=>a=b=c=1
