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

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

$\displaystyle (a+x)(b+y)(c+z)+4\left(\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}\right)\ge 20$.

My solution:

$\displaystyle (a+x)(b+y)(c+z)+4\left(\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}\right)$

$\displaystyle =abc+abz+bcx+acy+ayz+bxz+xyc+xyz+\frac{1}{ax}+\frac{1}{ax}+\frac{1}{ax}+\frac{1}{ax}$

$\displaystyle \,\,\,\,+\frac{1}{by}+\frac{1}{by}+\frac{1}{by}+\frac{1}{by}+\frac{1}{cz}+\frac{1}{cz}+\frac{1}{cz}+\frac{1}{cz}$

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

$\displaystyle = 20\sqrt[20]{1}$

$\displaystyle = 20$ (Q.E.D.)

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