Câu hỏi của Nguyễn Thị Thu Hằng

Question

Cho 3 số a,b,c thỏa mãn $a+b+c=\frac{3}{2}$ .Tìm gtnn của M=$a^2+b^2+c^2$

Nguyễn Việt Lâm · Accepted Answer

Ta luôn có: $\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0$

$\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2ac+2bc$

$\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2ac+2bc$

$\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2$

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

$\Rightarrow M=a^2+b^2+c^2\ge\frac{1}{3}\left(\frac{3}{2}\right)^2=\frac{3}{4}$

$\Rightarrow M_{min}=\frac{3}{4}$ khi $a=b=c=\frac{1}{2}$Câu trả lời: Ta luôn có: $\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0$

$\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2ac+2bc$

$\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2ac+2bc$

$\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2$

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

$\Rightarrow M=a^2+b^2+c^2\ge\frac{1}{3}\left(\frac{3}{2}\right)^2=\frac{3}{4}$

$\Rightarrow M_{min}=\frac{3}{4}$ khi $a=b=c=\frac{1}{2}$

Nguyễn Việt Lâm · Answer

Áp dụng BĐT:

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

$\Rightarrow M_{min}=\frac{3}{4}$ khi $a=b=c=\frac{1}{2}$Câu trả lời: Áp dụng BĐT:

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

$\Rightarrow M_{min}=\frac{3}{4}$ khi $a=b=c=\frac{1}{2}$

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