Question

Cho a, b, c >0 và a + b + c ≤ 3 . Chứng minh rằng :

\(\frac{4}{a^2+b^2+c^2}+\frac{2021}{ab++bc+ca}\) ≥ 675

Answer 1

\(\frac{4}{a^2+b^2+c^2}+\frac{2021}{ab+bc+ac}=\frac{4}{a^2+b^2+c^2}+\frac{4}{ab+bc+ac}+\frac{4}{ab+bc+ac}+\frac{2013}{ab+bc+ac}\)

\(=4\left(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}\right)+\frac{2013}{ab+bc+ac}\)

\(\ge\frac{36}{\left(a+b+c\right)^2}+\frac{2013}{ab+bc+ac}\ge\frac{36}{\left(a+b+c\right)^2}+\frac{2013}{\frac{\left(a+b+c\right)^2}{3}}\ge4+671=675\)

\("="\Leftrightarrow a=b=c=1\)

Answer 2

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