Cho a,b,c là số thực dương thỏa mãn $ 28\frac{(bc)^2 + (ac)^2+ (ab)^2}{(abc)^2} = 4(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}) + 2013 $ .Hãy tìm Max của :
$ P = \frac{1}{\sqrt{5a^2 + 2ab + b^2}} + \frac{1}{\sqrt{5b^2 + 2bc + c^2}} + \frac{1}{\sqrt{5c^2 + 2ac + c^2}} $
Điểm rơi xấu quá :<
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Theo BĐT AM-GM ta có:
[tex]\dfrac{1}{a^2}+\dfrac{1}{b^2}\geq \dfrac{2}{ab}[/tex]
Tương tụ..
Suy ra [tex]\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\leq \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}[/tex]
Ta có [tex]28\dfrac{(bc)^2 + (ac)^2+ (ab)^2}{(abc)^2} = 4(\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ac}) + 2013\\\Leftrightarrow 28\left ( \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \right )= 4(\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ac}) + 2013\\\Leftrightarrow 4(\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ac}) + 2013-28\left ( \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \right )=0[/tex]
Suy ra [tex]0=4(\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ac}) + 2013-28\left ( \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \right )\leq 4\left ( \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \right )+ 2013-28\left ( \dfrac{1}{a^2}+\frac{1}{b^2}+\dfrac{1}{c^2} \right )\\\Leftrightarrow 0\leq 2013-24\left ( \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \right )\\\Leftrightarrow \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \leq \dfrac{2013}{24}[/tex]
Ta có: [tex]5a^2 + 2ab + b^2=4a^2+4ab+(a-b)^2\geq 4a^2+4ab\Rightarrow \dfrac{1}{\sqrt{5a^2 + 2ab + b^2}}\leq \dfrac{1}{\sqrt{ 4a^2+4ab}}=\dfrac{1}{2\sqrt{a^2+ab}}[/tex]
Tương tự...
Suy ra [tex]P\leq \dfrac{1}{2}\left ( \dfrac{1}{\sqrt{a^2+ab}}+\dfrac{1}{\sqrt{b^2+bc}}+\dfrac{1}{\sqrt{c^2+ca}} \right )[/tex]
Theo BĐT BSC ta có:
[tex]\dfrac{1}{\sqrt{a^2+ab}}+\dfrac{1}{\sqrt{b^2+bc}}+\dfrac{1}{\sqrt{c^2+ca}}\leq \sqrt{3\left ( \dfrac{1}{a^2+ab}+\dfrac{1}{b^2+bc}+\dfrac{1}{c^2+ca} \right )}[/tex]
Ta có [tex]\dfrac{1}{a^2+ab}=\frac{1}{4}.\dfrac{4}{a^2+ab}\leq \dfrac{1}{4}\left ( \dfrac{1}{a^2}+\dfrac{1}{ab} \right )[/tex]
Tương tự....
Suy ra [tex]\dfrac{1}{a^2+ab}+\dfrac{1}{b^2+bc}+\dfrac{1}{c^2+ca}\\\leq \dfrac{1}{4}\left ( \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca} \right )\\\leq \dfrac{1}{4}\left ( \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \right )\\\leq \dfrac{1}{4}.2.\dfrac{2013}{24}=\dfrac{671}{16}[/tex]
Suy ra [tex]P\leq \dfrac{1}{2}.\sqrt{3.\dfrac{671}{16}}=\dfrac{1}{2}.\sqrt{\dfrac{2013}{16}}[/tex]
Dấu = xảy ra khi [tex]a=b=c=\sqrt{\dfrac{24}{671}}[/tex]
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