Question

Cho a;b;c thỏa mãn \(a+b+c=\frac{3}{2}\) Chứng minh rằng: \(a^2+b^2+c^2\ge\frac{3}{4}\)

Answer 1

Áp dụng Bunyakovsky , có :

\(\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2=\frac{9}{4}\)

\(\Rightarrow a^2+b^2+c^2\ge\frac{9}{4}.\frac{1}{3}=\frac{3}{4}\)

Đẳng thức xảy ra 

\(\Leftrightarrow a=b=c=\frac{1}{2}\)

Answer 2

Ta có :

a² + b² + c²

hay(a + b + c)² = \(\left(\frac{3}{2}\right)^2\)=\(\frac{6}{4}\)

Vậy a² + b² + c² >\(\frac{3}{4}\)

k mk nha

Answer 3

Ko cần dùng Cauchy dạng Engel luôn

Ta cần CM  \(a^2+b^2+c^2\ge ab+ac+bc\) (1)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(đúng)

Nên \(\left(1\right)\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac\ge3ab+3ac+3bc\)

\(\Leftrightarrow\left(a+b+c\right)^2\ge3ab+3ac+3bc\Leftrightarrow\frac{9}{4}\ge3\left(ab+ac+bc\right)\)

\(\Rightarrow ab+ac+bc\le\frac{3}{4}\)

Ta có : \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)=\frac{9}{4}\)

\(\Leftrightarrow a^2+b^2+c^2=\frac{9}{4}-2\left(ab+bc+ac\right)\)

\(\Rightarrow a^2+b^2+c^2\ge\frac{9}{4}-2.\frac{3}{4}=\frac{3}{4}\) (đpcm)