[62] - Algebraic Inequalities from and for Math Contests

Αν a,b,c>0 και a+b+c=1, τότε αποδείξτε ότι:

(a)

$$\frac{1 - a^2}{b + c} + \frac{1 - b^2}{c + a} + \frac{1 - c^2}{a + b} = 4$$

(b)

$$\frac{1-a^3}{b+c}+\frac{1-b^3}{c+a}+\frac{1-c^3}{a+b}\ge \frac{13}{3}$$