\(BĐT\Leftrightarrow35\left(a^2+b^2+c^2\right)\ge9\left(a+b+c\right)^2+\dfrac{72abc}{a+b+c}\)
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow35\left(a^2+b^2+c^2\right)\ge9\left(a+b+c\right)^2+8\left(a^2+b^2+c^2\right)\)
Cần chứng minh rằng \(8\left(a^2+b^2+c^2\right)\ge\dfrac{72abc}{a+b+c}\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)\left(a+b+c\right)\ge9abc\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\left(a^2+b^2+c^2\right)\left(a+b+c\right)\ge3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}=9abc\left(đpcm\right)\)
Vậy \(8\left(a^2+b^2+c^2\right)\ge\dfrac{72abc}{a+b+c}\) \(\Rightarrow9\left(a+b+c\right)^2+8\left(a^2+b^2+c^2\right)\ge9\left(a+b+c\right)^2+\dfrac{72abc}{a+b+c}\)
\(\Rightarrow\)\(35\left(a^2+b^2+c^2\right)\ge9\left(a+b+c\right)^2+\dfrac{72abc}{a+b+c}\) ( đpcm )