Câu hỏi của Nguyễn Ngọc Anh

Question

Chứng minh rằng với mọi a,b,c ta có a2+4b2+3c2>2a+12b+6c-14

Nguyễn Đức Trí · Accepted Answer

$\left(a-1\right)^2\ge0\Rightarrow a^2+1-2a\ge0\Rightarrow a^2+1\ge2a\left(1\right)$

$\left(2b-3\right)^2\ge0\Rightarrow4b^2+9-12b\ge0\Rightarrow4b^2+9\ge12b\left(2\right)$

$\left(c\sqrt[]{3}-\sqrt[]{3}\right)^2\ge0\Rightarrow3c^2+3-6c\ge0\Rightarrow3c^2+3\ge6c\left(3\right)$

$\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow a^2+1+4b^2+9+3c^2+3\ge2a+12b+6c$

$\Rightarrow a^2+4b^2+3c^2+1+9+3\ge2a+12b+6c$

$\Rightarrow a^2+4b^2+3c^2+13\ge2a+12b+6c$

$\Rightarrow a^2+4b^2+3c^2\ge2a+12b+6c-13$

mà $2a+12b+6c-13>2a+12b+6c-14$

$\Rightarrow a^2+4b^2+3c^2>2a+12b+6c-14$

$\Rightarrow dpcm$Câu trả lời: $\left(a-1\right)^2\ge0\Rightarrow a^2+1-2a\ge0\Rightarrow a^2+1\ge2a\left(1\right)$

$\left(2b-3\right)^2\ge0\Rightarrow4b^2+9-12b\ge0\Rightarrow4b^2+9\ge12b\left(2\right)$

$\left(c\sqrt[]{3}-\sqrt[]{3}\right)^2\ge0\Rightarrow3c^2+3-6c\ge0\Rightarrow3c^2+3\ge6c\left(3\right)$

$\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow a^2+1+4b^2+9+3c^2+3\ge2a+12b+6c$

$\Rightarrow a^2+4b^2+3c^2+1+9+3\ge2a+12b+6c$

$\Rightarrow a^2+4b^2+3c^2+13\ge2a+12b+6c$

$\Rightarrow a^2+4b^2+3c^2\ge2a+12b+6c-13$

mà $2a+12b+6c-13>2a+12b+6c-14$

$\Rightarrow a^2+4b^2+3c^2>2a+12b+6c-14$

$\Rightarrow dpcm$

Nguyễn Long Thành · Answer

⇔(a−1)2+(2b−3)2+3(c−1)2+1>0 (luôn đúng)

⇒⇒ BĐT ban đầu đúngCâu trả lời: ⇔(a−1)2+(2b−3)2+3(c−1)2+1>0 (luôn đúng)

⇒⇒ BĐT ban đầu đúng