Question

Giả thiết x, y, z > 0 và xy + y² + zx = a. Chứng minh rằng :
\(x\sqrt{\dfrac{\left(a+y^2\right)\left(a+z^2\right)}{a+x^2}}+y\sqrt{\dfrac{\left(a+z^2\right)\left(a+x^2\right)}{a+y^2}}+z\sqrt{\dfrac{\left(a+x^2\right)\left(a+y^2\right)}{a+z^2}}=2a\)

Answer 1

Ta có :

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

Do đó :

\(VT=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\left(x+z\right)\right)}}+y\sqrt{\dfrac{\left(x+z\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(x+z\right)\left(x+y\right)\left(y+z\right)}{\left(x+z\right)\left(y+z\right)}}\)

\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)

\(=2\left(xy+yz+zx\right)\)

\(=2a\) ( đpcm )