Bổ xung : ĐK : a;b;c > 0

Đặt \(A=\frac{a^3.c}{b}+\frac{a^3.b}{c}+\frac{b^3.c}{a}+\frac{b^3.a}{c}+\frac{c^3.a}{b}+\frac{c^3.b}{a}\)

Áp dụng AM - GM ta có :

\(A\ge6\sqrt[6]{\frac{a^3.c}{b}.\frac{a^3.b}{c}.\frac{b^3.c}{a}.\frac{b^3.a}{c}.\frac{c^3.a}{b}.\frac{c^3.b}{a}}=6\sqrt[6]{\left(abc\right)^6}=6abc\)(đpcm)

thanks nhìu lắm...... mà trang hk có lớp 10 nhỉ?

\(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)

\(=2\left(1+abc\right)+\sqrt{\left[\left(a+1\right)^2+\left(1-a\right)^2\right]\left[\left(b+c\right)^2+\left(bc-1\right)^2\right]}\)

\(\ge2\left(1+abc\right)+\left(a+1\right)\left(b+c\right)+\left(1-a\right)\left(bc-1\right)\)

\(=\left(1+a\right)\left(1+b\right)\left(1+c\right)\)

\(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}.\)

\(=2\left(1+abc\right)+\sqrt{\left[\left(a+1\right)^2+\left(1-a\right)^2\right]\left[\left(b+c\right)^2+\left(bc-1\right)^2\right]}\)

\(\ge2\left(1+abc\right)+\left(a+1\right)\left(b+c\right)+\left(1-a\right)\left(bc-1\right)\)

\(=\left(1+a\right)\left(1+b\right)\left(1+c\right)\)

\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2=1\)

Biến đổi vế trái ta có:

\(=\left[\frac{1-\sqrt{a^3}}{1-\sqrt{a}}+\sqrt{a}\right]\left[\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]^2\)

\(=\left[\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right]\left[\frac{1}{1+\sqrt{a}}\right]^2\)

\(=\left(1+\sqrt{a}+a+\sqrt{a}\right)\left(\frac{1}{a+2\sqrt{a}+1}\right)\)

\(=\frac{\left(a+2\sqrt{a}+1\right)}{a+2\sqrt{a}+1}\)

\(=1=VP\)

Vậy đẳng thức được chứng minh