\(\frac{1}{\left(1+a\right)^2}+\frac{1}{\left(1+b\right)^2}+\frac{1}{\left(1+b\right)^2}+\frac{2}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge1\)

<=> \(\left(1+b\right)^2\left(1+c\right)^2+\left(1+a\right)^2\left(1+b\right)^2+\left(1+a\right)\left(1+c\right)^2\)

\(+2\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+a\right)^2\left(1+b\right)^2\left(1+c\right)^2\)

<=> \(a^2+b^2+c^2\ge3\)đúng vì \(a^2+b^2+c^2\ge3\sqrt[3]{\left(abc\right)^2}=3\)

Dấu "=" xảy ra <=> a = b = c = 1

\(\left(\frac{x\sqrt{x}+1}{\sqrt{x}+1}-x\right)\left(\frac{x\sqrt{x}-1}{\sqrt{x}-1}-x\right)\)( đK: x\(\ge\)0 ; x\(\ne\)1) 

Ta có: \(x\sqrt{x}+1=\left(\sqrt{x}\right)^3+1=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)

\(x\sqrt{x}-1=\left(\sqrt{x}\right)^3-1=\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)

Do đó: 

\(\left(\frac{x\sqrt{x}+1}{\sqrt{x}+1}-x\right)\left(\frac{x\sqrt{x}-1}{\sqrt{x}-1}-x\right)\)

\(=\left(x-\sqrt{x}+1-x\right)\left(x+\sqrt{x}+1-x\right)\)

\(=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)=1-x\)

\(x_1+x_2=3+2\sqrt{3}+3-2\sqrt{3}=6\)

\(x_1.x_2=3^2-\left(2\sqrt{3}\right)^2=-3\)

=> Phương trình bậc 2 có dạng: x^2 - 6x - 3 = 0