Let $a,\,b,\,c,\,x,\,y$ and $z$ be strictly positive real numbers, prove that [MATH](a+x)(b+y)(c+z)+4\left(\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}\right)\ge 20[/MATH].

Let $a,\,b,\,c,\,x,\,y$ and $z$ be strictly positive real numbers, prove that

[MATH](a+x)(b+y)(c+z)+4\left(\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}\right)\ge 20[/MATH].

My solution:

[MATH](a+x)(b+y)(c+z)+4\left(\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}\right)[/MATH]

[MATH]=abc+abz+bcx+acy+ayz+bxz+xyc+xyz+\frac{1}{ax}+\frac{1}{ax}+\frac{1}{ax}+\frac{1}{ax}[/MATH]

[MATH]\,\,\,\,+\frac{1}{by}+\frac{1}{by}+\frac{1}{by}+\frac{1}{by}+\frac{1}{cz}+\frac{1}{cz}+\frac{1}{cz}+\frac{1}{cz}[/MATH]

[MATH]\ge 20\sqrt[20]{abc(abz)(bcx)(acy)(ayz)(bxz)(xyc)(xyz)\left(\frac{1}{ax}\right)^4\left(\frac{1}{by}\right)^4\left(\frac{1}{cz}\right)^4}[/MATH]

[MATH]= 20\sqrt[20]{1}[/MATH]

[MATH]= 20[/MATH] (Q.E.D.)

Equality occurs when $a=b=c=x=y=z=1.$