Question

Cho ba số thực dương a,b,c thỏa mãn abc = 1. Chứng minh rằng 

\(\left(a^2+b^2+c^2\right)^3\) ≥ 9(a + b + c)

Answer 1

 ≥ 9(a + b + c)

Answer 2

(a2+b2+c2)3(a2+b2+c2)3 ≥ 9(a + b + c)

Answer 3

chúc bn hok tốt

Answer 4

Ta có:

\(a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\)

\(\Rightarrow\left(a^2+b^2+c^2\right)^3\ge\left(\dfrac{1}{3}\left(a+b+c\right)^2\right)^3=\dfrac{1}{27}\left(a+b+c\right)^6\)

\(\Rightarrow VT\ge\dfrac{1}{27}\left(a+b+c\right)\left(a+b+c\right)^5\ge\dfrac{1}{27}\left(a+b+c\right)\left(3\sqrt[3]{abc}\right)^5=9\left(a+b+c\right)\) (đpcm)