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Câu hỏi: Với $a,b,c>0$ thỏa mãn $c=8ab$ thì biểu thức $P=\dfrac{1}{4a+2b+3}+\dfrac{c}{4bc+3c+2}+\dfrac{c}{2ac+3c+4}$ đạt giá trị lớn nhất bằng $\dfrac{m}{n}$ ( $m,n\in \mathbb{Z}$ và $\dfrac{m}{n}$ là phân số tối giản). Tính $2{{m}^{2}}+n$ ?
A. 9.
B. 4.
C. 8.
D. 3.
A. 9.
B. 4.
C. 8.
D. 3.
Ta có: $P=\dfrac{1}{4a+2b+3}+\dfrac{c}{4bc+3c+2}+\dfrac{c}{2a+3c+4}$
$=\dfrac{1}{4a+2b+3}+\dfrac{c}{4b+3+\dfrac{2}{c}}+\dfrac{c}{2a+3+\dfrac{4}{c}}$
Đặt $2a=x;2b=y;\dfrac{2}{c}=z\Rightarrow xyz=2a.2b.\dfrac{2}{c}=\dfrac{8ab}{c}=1$ (vì $c=8ab$ )
Khi đó ta có $P=\dfrac{1}{2x+y+3}+\dfrac{c}{2y+z+3}+\dfrac{c}{2z+x+3}$
Lại có: $2x+y+3=x+x+y+1+2\ge 2\sqrt{xy}+2\sqrt{x}+2=2\left( \sqrt{xy}+\sqrt{x}+1 \right)$
Tương tự có: $2y+z+3\le 2\left( \sqrt{yz}+\sqrt{y}+1 \right); 2z+x+3\le 2\left( \sqrt{xz}+\sqrt{z}+1 \right)$
Do đó ta có:
$P\le \dfrac{1}{2}\left( \dfrac{1}{\sqrt{xy}+\sqrt{x}+1}+\dfrac{1}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{1}{\sqrt{xz}+\sqrt{z}+1} \right)$
$=\dfrac{1}{2}\left( \dfrac{1}{\sqrt{xy}+\sqrt{x}+1}+\dfrac{1}{\dfrac{1}{\sqrt{x}}+\sqrt{y}+1}+\dfrac{1}{\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{xy}}+1} \right)$ (do $xyz=1$ )
$=\dfrac{1}{2}\left( \dfrac{1}{\sqrt{xy}+\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+1}+\dfrac{\sqrt{xy}}{\sqrt{xy}+\sqrt{x}+1} \right)$
$=\dfrac{1}{2}.\dfrac{\sqrt{xy}+\sqrt{x}+1}{\sqrt{xy}+\sqrt{x}+1}=\dfrac{1}{2}\Rightarrow P\le \dfrac{1}{2}$
Dấu "=" xảy ra khi $x=y=z=1$
Do đó: $\max P=\dfrac{1}{2}\Rightarrow m=1;n=2\Rightarrow 2{{m}^{2}}+n=4$
Note 92:
Bất đẳng thức Cô-si: Với 2 số x; y không âm ta có: $x+y\ge 2\sqrt{xy}$
Dấu " = " xảy ra $\Leftrightarrow x=y$
$=\dfrac{1}{4a+2b+3}+\dfrac{c}{4b+3+\dfrac{2}{c}}+\dfrac{c}{2a+3+\dfrac{4}{c}}$
Đặt $2a=x;2b=y;\dfrac{2}{c}=z\Rightarrow xyz=2a.2b.\dfrac{2}{c}=\dfrac{8ab}{c}=1$ (vì $c=8ab$ )
Khi đó ta có $P=\dfrac{1}{2x+y+3}+\dfrac{c}{2y+z+3}+\dfrac{c}{2z+x+3}$
Lại có: $2x+y+3=x+x+y+1+2\ge 2\sqrt{xy}+2\sqrt{x}+2=2\left( \sqrt{xy}+\sqrt{x}+1 \right)$
Tương tự có: $2y+z+3\le 2\left( \sqrt{yz}+\sqrt{y}+1 \right); 2z+x+3\le 2\left( \sqrt{xz}+\sqrt{z}+1 \right)$
Do đó ta có:
$P\le \dfrac{1}{2}\left( \dfrac{1}{\sqrt{xy}+\sqrt{x}+1}+\dfrac{1}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{1}{\sqrt{xz}+\sqrt{z}+1} \right)$
$=\dfrac{1}{2}\left( \dfrac{1}{\sqrt{xy}+\sqrt{x}+1}+\dfrac{1}{\dfrac{1}{\sqrt{x}}+\sqrt{y}+1}+\dfrac{1}{\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{xy}}+1} \right)$ (do $xyz=1$ )
$=\dfrac{1}{2}\left( \dfrac{1}{\sqrt{xy}+\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+1}+\dfrac{\sqrt{xy}}{\sqrt{xy}+\sqrt{x}+1} \right)$
$=\dfrac{1}{2}.\dfrac{\sqrt{xy}+\sqrt{x}+1}{\sqrt{xy}+\sqrt{x}+1}=\dfrac{1}{2}\Rightarrow P\le \dfrac{1}{2}$
Dấu "=" xảy ra khi $x=y=z=1$
Do đó: $\max P=\dfrac{1}{2}\Rightarrow m=1;n=2\Rightarrow 2{{m}^{2}}+n=4$
Note 92:
Bất đẳng thức Cô-si: Với 2 số x; y không âm ta có: $x+y\ge 2\sqrt{xy}$
Dấu " = " xảy ra $\Leftrightarrow x=y$
Đáp án C.