Toán cho a,b,c lớn hơn bằng không cm a+b+c lớn hơn bằng căn ab+căn bc+ căn ac 16/07/2021 By Lyla cho a,b,c lớn hơn bằng không cm a+b+c lớn hơn bằng căn ab+căn bc+ căn ac
Lời giải chi tiết: \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\) \(\Rightarrow a+b+b+c+c+a\ge2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}\) \(\Rightarrow a+b+b+c+c+a-2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ac}\ge0\) \(\Rightarrow\left(a+b-2\sqrt{ab}\right)+\left(b+c-2\sqrt{bc}\right)+\left(c+a-2\sqrt{ac}\right)\ge0\) \(\Rightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2\ge0\) luôn đúng Dấu `’=’` xảy ra khi `a=b=c` Trả lời
$a,\,b,\,c\ge 0$ Ta có: $\begin{cases}(\sqrt{a}-\sqrt{b})^2\ge 0\\(\sqrt{b}-\sqrt{c})^2\ge 0\\(\sqrt{a}-\sqrt{c})^2\ge 0\end{cases}$ $⇔(\sqrt{a}-\sqrt{b})^2+(\sqrt{b}-\sqrt{c})^2+(\sqrt{a}-\sqrt{c})^2\ge 0$ $⇔a-2\sqrt{ab}+b+b-2\sqrt{bc}+c+a-2\sqrt{ac}+c\ge 0$ $⇔2(a+b+c)-2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})\ge 0$ $⇔2(a+b+c)\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$ $⇔a+b+c\ge \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$ (Đpcm). Trả lời
Lời giải chi tiết:
\(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)
\(\Rightarrow a+b+b+c+c+a\ge2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}\) \(\Rightarrow a+b+b+c+c+a-2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ac}\ge0\) \(\Rightarrow\left(a+b-2\sqrt{ab}\right)+\left(b+c-2\sqrt{bc}\right)+\left(c+a-2\sqrt{ac}\right)\ge0\) \(\Rightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2\ge0\) luôn đúng
Dấu `’=’` xảy ra khi `a=b=c`
$a,\,b,\,c\ge 0$
Ta có: $\begin{cases}(\sqrt{a}-\sqrt{b})^2\ge 0\\(\sqrt{b}-\sqrt{c})^2\ge 0\\(\sqrt{a}-\sqrt{c})^2\ge 0\end{cases}$
$⇔(\sqrt{a}-\sqrt{b})^2+(\sqrt{b}-\sqrt{c})^2+(\sqrt{a}-\sqrt{c})^2\ge 0$
$⇔a-2\sqrt{ab}+b+b-2\sqrt{bc}+c+a-2\sqrt{ac}+c\ge 0$
$⇔2(a+b+c)-2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})\ge 0$
$⇔2(a+b+c)\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$
$⇔a+b+c\ge \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$ (Đpcm).