Câu hỏi của Ánh Dương

Question

1.Cho các số a, b, c $\in\left[0;1\right]$. Cmr: $a+b^2+c^3-ab-bc-ca\le1$

2. Cho x>0, y>0 và $x+y\ge6$. Tìm min của $P=3x+2y+\frac{6}{x}+\frac{8}{y}$

Nguyễn Việt Lâm · Accepted Answer

2.

$P=3x+2y+\frac{6}{x}+\frac{8}{y}$

$P=\frac{3x}{2}+\frac{6}{x}+\frac{y}{2}+\frac{8}{y}+\frac{3x}{2}+\frac{3y}{2}$

$P=\left(\frac{3x}{2}+\frac{6}{x}\right)+\left(\frac{y}{2}+\frac{8}{y}\right)+\frac{3}{2}\left(x+y\right)$

$P\ge2\sqrt{\frac{18x}{2x}}+2\sqrt{\frac{8y}{2y}}+\frac{3}{2}.6=19$

$P_{min}=19$ khi $\left\{{}\begin{matrix}x=2\y=4\end{matrix}\right.$Câu trả lời: 2.

$P=3x+2y+\frac{6}{x}+\frac{8}{y}$

$P=\frac{3x}{2}+\frac{6}{x}+\frac{y}{2}+\frac{8}{y}+\frac{3x}{2}+\frac{3y}{2}$

$P=\left(\frac{3x}{2}+\frac{6}{x}\right)+\left(\frac{y}{2}+\frac{8}{y}\right)+\frac{3}{2}\left(x+y\right)$

$P\ge2\sqrt{\frac{18x}{2x}}+2\sqrt{\frac{8y}{2y}}+\frac{3}{2}.6=19$

$P_{min}=19$ khi $\left\{{}\begin{matrix}x=2\y=4\end{matrix}\right.$

Nguyễn Việt Lâm · Answer

1.

Do $0\le a;b;c\le1\Rightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge0$

$\Leftrightarrow1-abc-a-b-c+ab+bc+ca\ge0$

$\Leftrightarrow a+b+c-ab-bc-ca\le1-abc\le1$

Mặt khác $0\le a;b;c\le1\Rightarrow\left\{{}\begin{matrix}b^2\le b\c^3\le c\end{matrix}\right.$

$\Rightarrow a+b^2+c^3-ab-bc-ca\le a+b+c-ab-bc-ca\le1$ (đpcm)

Dấu "=" xảy ra khi $\left(a;b;c\right)=\left(0;0;1\right)$ và hoán vịCâu trả lời: 1.

Do $0\le a;b;c\le1\Rightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge0$

$\Leftrightarrow1-abc-a-b-c+ab+bc+ca\ge0$

$\Leftrightarrow a+b+c-ab-bc-ca\le1-abc\le1$

Mặt khác $0\le a;b;c\le1\Rightarrow\left\{{}\begin{matrix}b^2\le b\c^3\le c\end{matrix}\right.$

$\Rightarrow a+b^2+c^3-ab-bc-ca\le a+b+c-ab-bc-ca\le1$ (đpcm)

Dấu "=" xảy ra khi $\left(a;b;c\right)=\left(0;0;1\right)$ và hoán vị