Áp dụng bất đẳng thức Bunyakovsky ta có:
\(\dfrac{1}{a}+\dfrac{4}{b}+\dfrac{9}{c}=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{4}{b}+\dfrac{9}{c}\right)\)
\(=\left(\sqrt{a}^2+\sqrt{b}^2+\sqrt{c}^2\right)\left[\left(\dfrac{1}{\sqrt{a}}\right)^2+\left(\dfrac{2}{\sqrt{b}}\right)^2+\left(\dfrac{3}{\sqrt{c}}\right)^2\right]\)
\(\ge\left(\sqrt{a}.\dfrac{1}{\sqrt{a}}+\sqrt{b}.\dfrac{2}{\sqrt{b}}+\sqrt{c}.\dfrac{3}{\sqrt{c}}\right)^2\)
\(=\left(1+2+3\right)^2=36\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}a=\dfrac{1}{6}\\b=\dfrac{1}{3}\\c=\dfrac{1}{2}\end{matrix}\right.\)
Vì đề không cho a\(\ne\) b\(\ne\)c
Theo đề ta có:
\(\dfrac{1}{a}+\dfrac{4}{b}+\dfrac{9}{c}\)=> \(\dfrac{1+4+9}{a+b+c}=\dfrac{14}{1}=14\)
Mà 14> 36
Nên \(\dfrac{1}{a}+\dfrac{4}{b}+\dfrac{9}{c}\) =14 > 36 ( đpcm)
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