Question

Cho a,b,c là 3 cạnh của một tam giác thỏa \(a+b+c=2\)
Chứng minh \(a^2+b^2+c^2\ge\frac{52}{27}\)

Answer 1

Chỉnh sửa: \(a^2+b^2+c^2+2abc\ge\frac{52}{27}\)

Answer 2

Theo BĐT AM-GM ta có:

\(\left(1-a\right)\left(1-b\right)\left(1-c\right)\le\left(\frac{3-\left(a+b+c\right)}{3}\right)^3=\frac{1}{27}\)

\(\Leftrightarrow ab+bc+ca+1-\left(a+b+c\right)-abc\le\frac{1}{27}\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)-\left(a^2+b^2+c^2+2abc\right)\le\frac{56}{27}\)

\(\Leftrightarrow\left(a+b+c\right)^2-\left(a^2+b^2+c^2+2abc\right)\le\frac{56}{27}\)

\(\Leftrightarrow4-\left(a^2+b^2+c^2+2abc\right)\le\frac{56}{27}\)

\(\Leftrightarrow a^2+b^2+c^2+2abc\ge\frac{52}{27}\)