Question

[Trần thị vân]

Cho các số a,b,c thỏa mãn a+b+c=\(\dfrac{3}{2}\) chứng minh \(a^2+b^2+c^2\ge\dfrac{3}{4}\)

Answer 1

[Huong San]

Ta có:(A1)\(^2\)\(\ge\)0

\(\Leftrightarrow a^2-a+\dfrac{1}{4}\ge0\\ \Leftrightarrow a^2+\dfrac{1}{4}\ge a\left(1\right)\\ cmtt:b^2+\dfrac{1}{4}\ge b\left(2\right)\\ 6^2+\dfrac{1}{4}\ge c\left(3\right)\)

Cộng (1);(2) và (3) theo vế, ta có:

\(a^2+\dfrac{1}{4}+b^2+\dfrac{1}{4}+6^2+\dfrac{1}{4}\ge a+b+c\\ \Leftrightarrow a^2+b^2+c^2+\dfrac{3}{4}\ge\dfrac{3}{2}\\ \Leftrightarrow a^2+b^2+c^2\ge\dfrac{3}{2}-\dfrac{3}{4}\\ \Leftrightarrow a^2+b^2+c^2\ge\dfrac{3}{4}\)

Answer 2

[Phạm Tuấn Đạt]

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

\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=\dfrac{9}{4}\)

Có \(a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)

\(b^2+c^2\ge2\sqrt{b^2c^2}=2bc\)

\(a^2+c^2\ge2\sqrt{a^2c^2}=2ac\)

\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc\le a^2+b^2+c^2+a^2+b^2+a^2+c^2+b^2+c^2=3\left(a^2+b^2+c^2\right)\)

\(\Rightarrow\dfrac{9}{4}\le3\left(a^2+b^2+c^2\right)\)

\(\Rightarrow a^2+b^2+c^2\ge\dfrac{9}{4}.\dfrac{1}{3}=\dfrac{3}{4}\left(ĐPCM\right)\)

Bài này áp dụng BĐT cosi nha bn