Question

Using the relation $A.M\ge G.M$, prove that
(i) $(x^{2}y+y^{2}z+z^{2}x)(x{y}^2+y{z}^2+z{x}^2)\ge 9x^2{y}^2{z}^2$. ($x,y,z$ are positive real numbers)
(ii) $(a+b).(b+c).(c+a)>abc$; if $a,b,c$ are positive real numbers

Answer 1

$\left(i\right)\frac{(x^{2}y+y^{2}z+z^{2}x)(x{y}^2+y{z}^2+z{x}^2)}{9}=\frac{x^3{y}^3+3x^2{y}^2{z}^2+x^{4}yz+y^{4}xz+y^3{z}^3+z^{4}yx+z^3{x}^3}{9}=A.M$

$\sqrt[9]{9x^3{y}^3\times (x^2{y}^2{z}^2{)}^3\times x^{4}yz\times y^{4}xz\times z^{4}yx\times y^3{z}^3\times z^3{x}^3}=x^2{y}^2{z}^2=G.M$

As $A.M\ge G.M$,

$\frac{(x^{2}y+y^{2}z+z^{2}x)(x{y}^2+y{z}^2+z{x}^2)}{9}\ge x^2{y}^2{z}^2$

$\Rightarrow (x^{2}y+y^{2}z+z^{2}x)(x{y}^2+y{z}^2+z{x}^2)\ge 9x^2{y}^2{z}^2$

$\left(ii\right)\frac{(a+b)(b+c)(c+a)}{8}=\frac{(ab+ac+b^2+bc)(c+a)}{8}=\frac{2abc+a{c}^2+b^{2}c+b{c}^2+a^{2}b+a^{2}c+b^{2}a}{8}=A.M$

$\sqrt[8]{8(abc{)}^2\times a{c}^2\times a^{2}c\times b^{2}c\times b{c}^2\times a^{2}b\times a{b}^2}=abc=G.M$

As $A.M\ge G.M$,

$\frac{(a+b)(b+c)(c+a)}{8}\ge abc$

$\Rightarrow (a+b)(b+c)(c+a)\ge 8abc>abc$