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:
\(a^2+\dfrac{1}{4}+b^2+\dfrac{1}{4}+6^2+\dfrac{1}{4}\ge a+b+c\\ \Leftrightarrow a^2+b^2+c^2+\dfrac{3}{4}\ge\dfrac{3}{2}\\ \Leftrightarrow a^2+b^2+c^2\ge\dfrac{3}{2}-\dfrac{3}{4}\\ \Leftrightarrow a^2+b^2+c^2\ge\dfrac{3}{4}\)
\(\left(a+b+c\right)^2=\dfrac{9}{4}\)
\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=\dfrac{9}{4}\)
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
Cách khác : C1 : Áp dụng BĐT Bunhiacopxki vào bài toán , ta có :
\(\left(a^2+b^2+c^2\right)\left(1^2+1^2+1^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow a^2+b^2+c^2\ge\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{9}{4}.\dfrac{1}{3}=\dfrac{3}{4}\)
\("="\Leftrightarrow a=b=c=\dfrac{1}{2}\)
C2 : Áp dụng BĐT : \(\left(x-y\right)^2\ge0\forall xy\)
\(\Leftrightarrow x^2+y^2\ge2xy\)
\(\Rightarrow\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ac\end{matrix}\right.\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow a^2+b^2+c^2\ge\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{9}{4}.\dfrac{1}{3}=\dfrac{3}{4}\)
\("="\Leftrightarrow a=b=c=\dfrac{1}{2}\)
P/s : Không dùng đc Cauchy nhé vì đề chưa cho dương :3
Thêm cách nữa nè :3
\(Cauchy-Schwarz\): \(a^2+b^2+c^2\ge\dfrac{\left(a+b+c\right)^2}{1+1+1}=\dfrac{\left(\dfrac{3}{2}\right)^2}{3}=\dfrac{3}{4}\)
\("="\Leftrightarrow a=b=c\)
\(Cauchy-Schwarz:\)\(a^2+b^2+c^2\ge\dfrac{\left(a+b+c\right)^2}{1+1+1}=\dfrac{\left(\dfrac{3}{2}\right)^2}{3}=\dfrac{3}{4}\)
\("="\Leftrightarrow a=b=c=\dfrac{1}{2}\)
