Prove that $(b + c) (c + a) (a + b) > 8 a b c$

Open in App

Solution

This is true only if $a, b, c \geq 0$
Apply $A M \geq G M$ for $a, b$
$\frac{a + b}{2} \geq \sqrt{a b} \to$ eqn 1
Apply $A M \geq G M$ for $b, c$
$\frac{b + c}{2} \geq \sqrt{b c} \to$ eqn 2
Apply $A M \geq G M$ for $c, a$
$\frac{c + a}{2} \geq \sqrt{c a} \to$ eqn 3
Multiply corresponding sides of eqn 1,2,3
$\Rightarrow (a + b) (b + c) (c + a) \geq 8 a b c$

Suggest Corrections

Similar questions

(b + c) (c + a) (c + b) \geq 8 a b c

∣ ∣ ∣ ∣ \begin{matrix} b + c & c + a & a + b c + a & a + b & b + c a + b & b + c & c + a \end{matrix} ∣ ∣ ∣ ∣ = 2 ∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{3} & b^{3} & c^{3} \end{matrix} ∣ ∣ ∣ ∣ = (b - c) (c - a) (a - b) (a + b + c)

∣ ∣ ∣ ∣ \begin{matrix} b + c & c + a & a + b c + a & a + b & b + c a + b & b + c & c + a \end{matrix} ∣ ∣ ∣ ∣ == 2 (a + b + c) (a b + b c + c a - a^{2} - b^{2} - c^{2})

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{3} & c^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣

(a - b) (b - c) (c - a) (a + b + c) .

Join BYJU'S Learning Program