Question

Cho a+b+c=1. CMR : \(a^2+b^2+c^2+2\sqrt{3abc}\le1\)

Answer 1

Ta cần chứng minh

\(\left(a+b+c\right)^2\ge a^2+b^2+c^2+2\sqrt{3\left(a+b+c\right)abc}\)

\(\Leftrightarrow ab+bc+ca\ge\sqrt{3\left(a+b+c\right)abc}\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2-abc^2-bca^2-cab^2\ge0\)

\(\Leftrightarrow\left(ab-bc\right)^2+\left(bc-ca\right)^2+\left(ca-ab\right)^2\ge0\) (đúng)