Question

Với các số thực dương a, b, c thỏa mãn a² + b² + c² = 3. Chứng minh rằng: \(\Sigma\dfrac{1}{a}\ge\Sigma\dfrac{8}{a^2+7}\)

Answer 1

\(3=a^2+b^2+c^2\ge\dfrac{\left(a+b+c\right)^2}{3}\)

\(\Rightarrow a+b+c\le3\)

- Áp dụng bất đẳng thức Caushy-Schwarz, ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\ge\dfrac{9}{3}=3\)

\(\Rightarrow\sum\dfrac{1}{a}\ge3\)

\(\sum\dfrac{8}{a^2+7}=\sum\dfrac{8}{a^2+1+1+1+1+1+1+1}\le\sum\dfrac{8}{8.\sqrt[8]{a^2}}=\sum\sqrt[8]{\dfrac{1}{a}.\dfrac{1}{a}.1.1.1.1.1.1}\le\sum\dfrac{\dfrac{1}{a}+\dfrac{1}{a}+1+1+1+1+1+1}{8}=\sum\left(\dfrac{1}{4a}+\dfrac{3}{4}\right)=\sum\dfrac{1}{4a}+\dfrac{9}{4}\le\sum\dfrac{1}{4a}+\sum\dfrac{3}{4a}=\sum\dfrac{1}{a}\left(đpcm\right)\)

- Dấu "=" xảy ra khi \(a=b=c=1\)