Question 29
Let $a$, $b$, $c$ and $d$ be positive real numbers, all larger than or equal to $1$. Prove that
a. $abcd(a+b+c+d-4)^4 \ge (a-1)(b-1)(c-1)(d-1)(a+b+c+d)^4$,
b. $\dfrac{d}{a+b+c} + \dfrac{a}{b+c+d} + \dfrac{b}{c+d+a} + \dfrac{c}{d+a+b} \ge \dfrac{4}{3}$.
a. $abcd(a+b+c+d-4)^4 \ge (a-1)(b-1)(c-1)(d-1)(a+b+c+d)^4$,
b. $\dfrac{d}{a+b+c} + \dfrac{a}{b+c+d} + \dfrac{b}{c+d+a} + \dfrac{c}{d+a+b} \ge \dfrac{4}{3}$.