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Question: If \[x,y,z\] are positive real numbers, prove that \[{\left( {x + y + z} \right)^2}{\left( {yz + zx ...

If x,y,zx,y,z are positive real numbers, prove that (x+y+z)2(yz+zx+xy)2⩽3(y2+yz+z2)(z2+zx+x2)(x2+xy+y2){\left( {x + y + z} \right)^2}{\left( {yz + zx + xy} \right)^2} \leqslant 3\left( {{y^2} + yz + {z^2}} \right)\left( {{z^2} + zx + {x^2}} \right)\left( {{x^2} + xy + {y^2}} \right) .

Explanation

Solution

In order to solve the question given above, we will use the concept of arithmetic mean and geometric mean inequalities. The AM-GM inequalities states that the arithmetic mean of any set of non-negative real numbers is greater than or equal to the geometric mean of that set. Perform the question in a neat manner and in a step by step format to avoid any errors.

Complete step by step solution:
We have to prove (x+y+z)2(yz+zx+xy)2⩽3(y2+yz+z2)(z2+zx+x2)(x2+xy+y2){\left( {x + y + z} \right)^2}{\left( {yz + zx + xy} \right)^2} \leqslant 3\left( {{y^2} + yz + {z^2}} \right)\left( {{z^2} + zx + {x^2}} \right)\left( {{x^2} + xy + {y^2}} \right) .
We will begin our answer with the observation that:
x2+xy+y2=34(x+y)2+14(x−y)2⩾34(x+y)2{x^2} + xy + {y^2} = \dfrac{3}{4}{\left( {x + y} \right)^2} + \dfrac{1}{4}{\left( {x - y} \right)^2} \geqslant \dfrac{3}{4}{\left( {x + y} \right)^2}
In the same way the result will be the same fory2+yz+z2{y^2} + yz + {z^2} and for z2+zx+x2{z^2} + zx + {x^2} .
In the next step, we will be multiplying the values of all three, that is, x2+xy+y2{x^2} + xy + {y^2} , y2+yz+z2{y^2} + yz + {z^2} andz2+zx+x2{z^2} + zx + {x^2} .
On multiplying the values, we get,
3(y2+yz+z2)(z2+zx+x2)(x2+xy+y2)⩾8164(x+y)2(y+z)2(z+x)23\left( {{y^2} + yz + {z^2}} \right)\left( {{z^2} + zx + {x^2}} \right)\left( {{x^2} + xy + {y^2}} \right) \geqslant \dfrac{{81}}{{64}}{\left( {x + y} \right)^2}{\left( {y + z} \right)^2}{\left( {z + x} \right)^2} .
Now, with the help of this equation, we can prove that,
(x+y+z)(xy+yz+zx)⩽98(x+y)(y+z)(z+x)\left( {x + y + z} \right)\left( {xy + yz + zx} \right) \leqslant \dfrac{9}{8}\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right) .
This can be written as:
8(x+y+z)(xy+yz+zx)⩽9(x+y)(y+z)(z+x)8\left( {x + y + z} \right)\left( {xy + yz + zx} \right) \leqslant 9\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right) .
Also, (x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)−xyz\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right) = \left( {x + y + z} \right)\left( {xy + yz + zx} \right) - xyz .
From the above equation, we get that:
(x+y)(y+z)(z+x)⩾8xyz\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right) \geqslant 8xyz .
Now, this follows from the arithmetic mean and geometric mean inequalities: x+y⩾2xy,y+z⩾2yz,z+x⩾2zxx + y \geqslant 2\sqrt {xy} ,y + z \geqslant 2\sqrt {yz} ,z + x \geqslant 2\sqrt {zx} .

Note: To solve the sums similar to the one given above, always make sure you write the answer in a neat step by step form. To solve these questions, you need to remember the concept of arithmetic mean and geometric mean inequalities also popularly called as AM-GM inequalities. As mentioned above, the arithmetic and geometric mean inequalities that AM of any set of positive real numbers is greater than or equal to the GM of the same set.