Question

If x, y, z are real and positive, find the minimum value of

$\frac{( {x}^{2} + x + 1)( {y}^{2} + y + 1)( {z}^{2} + z + 1) }{xyz}$
​

Answer 1

$\red{\bigstar}$A.M-G.M inequality.

$\green{\bigstar}$The arithmetic mean(for short, A.M) is related to the value of the geometric mean(G.M). The A.M is always greater or equal to G.M which can be used in proofs of inequality.

$\implies\text{(A.M)}\geq\text{(G.M)}\text{ for positive numbers or 0}$

$\green{\bigstar}$(In use) A.M-G.M inequality.

Let's split the product into three numbers,

$\iff A=\dfrac{x^{2}+x+1}{x}=x+1+\dfrac{1}{x}$

$\iff B=\dfrac{y^{2}+y+1}{y}=y+1+\dfrac{1}{y}$

$\iff C=\dfrac{z^{2}+z+1}{z}=z+1+\dfrac{1}{z}$

to utilize the inequality.

$\iff A=x+\dfrac{1}{x}+1\geq2+1=3\text{ equals when }x=1$

$\iff B=y+\dfrac{1}{y}+1\geq2+1=3\text{ equals when }y=1$

$\iff C=z+\dfrac{1}{z}+1\geq2+1=3\text{ equals when }z=1$

Hence,

$\iff ABC\geq27\text{ equals when }x=y=z=1$

which leads to a minimum value, 27.

Answer 2

Answer:

Step-by-step explanation:

topic :

fraction

given :

• If x, y, z are real and positive, find the minimum value of

• $\frac{( {x}^{2} + x + 1)( {y}^{2} + y + 1)( {z}^{2} + z + 1) }{xyz}$

to find :

• [tex] \frac{( {x}^{2} + x + 1)( {y}^{2} + y + 1)( {z}^{2} + z + 1) }{xyz} [/tex

solution :

• please check the attached file please

• So real, positive , minimum value = 27