Question

Cho a , b , c là các số dương . Tìm GTNN :

A = \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

Answer 1

Áp dụng BĐT :

( x + y + z) \(\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) ≥ 9 ( x , y , z dương )

Trong đó : x = b + c ; y = a + c ; z = a + b

⇒ ( b + c + a + c + a + b)\(\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)\) ≥ 9 ( a > 0 ; b > 0 ; c > 0 )

⇔ 2( a + b + c)\(\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)\) ≥ 9

⇔\(\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}\) ≥ 4,5

⇔ 1 + \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+1+\dfrac{c}{a+b}+1\) ≥ 4,5

⇔\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\) ≥ 1,5

⇒ A_min = 1,5 ⇔ a = b = c