Question

Cho x, y, z > 0 và xy + yz + zx = a.

Chứng minh: \(x\sqrt{\frac{\left(a+y^2\right)\left(a+z^2\right)}{a+x^2}}+y\sqrt{\frac{\left(a+z^2\right)\left(a+x^2\right)}{a+y^2}}+z\sqrt{\frac{\left(a+x^2\right)\left(a+y^2\right)}{a+z^2}}=2a\)

Answer 1

Lời giải:

\(xy+yz+xz=a\Rightarrow \left\{\begin{matrix} x^2+a=x^2+xy+yz+xz=(x+y)(x+z)\\ y^2+a=y^2+xy+yz+xz=(y+x)(y+z)\\ z^2+a=z^2+xy+yz+xz=(z+x)(z+y)\end{matrix}\right.\)

Khi đó:
\(x\sqrt{\frac{(a+y^2)(a+z^2)}{a+x^2}}+y\sqrt{\frac{(a+z^2)(a+x^2)}{a+y^2}}+z\sqrt{\frac{(a+x^2)(a+y^2)}{a+z^2}}\)

\(=x\sqrt{\frac{(y+x)(y+z)(z+x)(z+y)}{(x+y)(x+z)}}+y\sqrt{\frac{(z+x)(z+y)(x+y)(x+z)}{(y+x)(y+z)}}+z\sqrt{\frac{(x+y)(x+z)(y+x)(y+z)}{(z+x)(z+y)}}\)

\(=x(y+z)+y(x+z)+z(x+y)=2(xy+yz+xz)=2a\)

Ta có đpcm.