- Áp dụng BĐT cauchuy : \(\left\{{}\begin{matrix}9m^2+n^2\ge2\sqrt{9m^2n^2}=6mn\\\dfrac{1}{9m^2}+\dfrac{1}{n^2}\ge2\sqrt{\dfrac{1}{9m^2n^2}}=\dfrac{2}{3mn}\end{matrix}\right.\)
\(\Rightarrow\left(9m^2+n^2\right)\left(\dfrac{1}{9m^2}+\dfrac{1}{n^2}\right)\ge6mn.\dfrac{2}{3mn}=4\left(1\right)\)
- Dấu " = " xảy ra <=> \(9m^2=n^2\)\(\Leftrightarrow\left(3m-n\right)\left(3m+n\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}n=3m\\n=-3m\end{matrix}\right.\)
Mà m, n > 0
\(\Rightarrow n=3m\)
Là 9m2 + n2.(\(\dfrac{1}{9m^2}+\dfrac{1}{n^2}\)) hay là (9m2 + n2)(\(\dfrac{1}{9m^2}+\dfrac{1}{n^2}\)) ?
A nghĩ đề là (9m2 + n2)(\(\dfrac{1}{9m^2}+\dfrac{1}{n^2}\)) \(\ge\) 4
Cách khác:
Ta có BĐT phụ \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) (a; b dương)
CM: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
\(\Leftrightarrow\) \(\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)
\(\Leftrightarrow\) (a + b)2 \(\ge\) 4ab
\(\Leftrightarrow\) (a + b)2 - 4ab \(\ge\) 0
\(\Leftrightarrow\) (a - b)2 \(\ge\) 0 (luôn đúng)
Dấu "=" xảy ra \(\Leftrightarrow\) a = b
Áp dụng BĐT phụ trên cho 2 số \(\dfrac{1}{9m^2}\) và \(\dfrac{1}{n^2}\) dương ta được:
\(\dfrac{1}{9m^2}+\dfrac{1}{n^2}\ge\dfrac{4}{9m^2+n^2}\)
\(\Leftrightarrow\) (9m2 + n2)(\(\dfrac{1}{9m^2}+\dfrac{1}{n^2}\)) \(\ge\) \(\dfrac{4\left(9m^2+n^2\right)}{9m^2+n^2}\) = 4 (đpcm)Dấu "=" xảy ra \(\Leftrightarrow\) 9m2 = n2
\(\Leftrightarrow\) n = 3m (do m, n > 0)
\(\Leftrightarrow\) m = \(\dfrac{n}{3}\)
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