Question

  giải bài 4 giúp em

Answer 1

Bài 5: .

a) Vì \(a,b,c\ge0\) nên ta có:

\(K=ab+4ac-4bc\ge-4bc\left(1\right)\)

Ta có: \(a+b+2c=1\Rightarrow b+2c=1-a\le1\) (do \(a\ge0\))

\(\Rightarrow b+2c\le1\left(2\right)\)

Theo Cauchy cho 2 số không âm ta có:

\(b+2c\ge2\sqrt{b.2c}\left(3\right)\)

Từ \(\left(2\right),\left(3\right)\Rightarrow2\sqrt{b.2c}\le1\)

\(\Rightarrow4.b.2c\le1\)

\(\Rightarrow4bc\le\dfrac{1}{2}\Rightarrow-4bc\ge-\dfrac{1}{2}\left(4\right)\)

Từ \(\left(1\right),\left(4\right)\Rightarrow K\ge-\dfrac{1}{2}\) 

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=0\\a+b+2c=1\\b=2c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0\\b=\dfrac{1}{2}\\c=\dfrac{1}{4}\end{matrix}\right.\)

b) Vì \(a,b,c\ge0\Rightarrow K=ab+4ac-4bc\le ab+4ac\left(1'\right)\)

\(a+b+2c=1\Rightarrow a^2+ab+2ac=a\)

\(\Rightarrow ab+4ac=a-2ac-a^2\le a-a^2\) (do \(ac\ge0\))

\(\Rightarrow ab+4ac\le a-a^2\)

Ta có \(a-a^2=-\left(a^2-a\right)=\left(a^2-a+\dfrac{1}{4}-\dfrac{1}{4}\right)=-\left(a-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

\(\Rightarrow ab+4ac\le\dfrac{1}{4}\left(2'\right)\)

Từ \(\left(1'\right),\left(2'\right)\Rightarrow K\le\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}-4bc=-2ac=0\\a+b+2c=1\\a-\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b=\dfrac{1}{2}\\c=0\end{matrix}\right.\)

Vậy \(MaxK=\dfrac{1}{4}\), đạt tại \(a=b=\dfrac{1}{2};c=0\)