Question

If a, b, c are real numbers then prove that (a+b) (b+c) (c+a) >=8abc

Answer 1

(a+b) (b+c) (c+a) >=8abc
=> a+b X b+c X c+a >=8abc
=> a x a x b x b x c x c >=8abc
=> (a^2) (b^2) (c^2) > 8abc

therefore, (a+b) (b+c) (c+a) > 8abc

Answer 2

Applying AM - GM inequality for a,b

$\qquad \displaystyle \sf \frac{a + b}{2} \: \geqslant \: b$

$\qquad \displaystyle \sf a + b \: \geqslant \: 2 \sqrt{ab}$

Similarly,

$\qquad \displaystyle \sf b + c \: \geqslant \: 2 \sqrt{bc}$

$\qquad \qquad \&$

$\qquad \displaystyle \sf c + a \: \geqslant \: 2 \sqrt{ca}$

$\sf\therefore \: (a + b)(b + c)(c + a) \: \geqslant \: 8 \sqrt{ab} \sqrt{bc} \sqrt{ca}$

$\qquad \qquad = \: \sf 8 \sqrt{(ab)(bc)(ca)}$

$\qquad \qquad = \: \sf 8 \sqrt{(abc)^{2} }$

$\qquad \qquad = \: \bf 8 \: abc$

Hence,

=> (a+b) (b+c) (c+a) = 8 abc

$\:$