Chương I - Căn bậc hai. Căn bậc ba

TS

Cho: \(4a^2+b^2+c^2\le4\). CMR: \(ab+bc+ca\le1+\sqrt{3}\)

H24
20 tháng 8 2018 lúc 20:50

Đây nhé @Liana

\(2a^2+\left(2-\sqrt{3}\right)b^2+2a^2+\left(2-\sqrt{3}\right)c^2+\left(\sqrt{3}-1\right)b^2+\left(\sqrt{3}-1\right)c^2\)

\(\ge2\sqrt{4-2\sqrt{3}}ab+2\sqrt{4-2\sqrt{3}}ac+2\left(\sqrt{3}-1\right)bc\)

\(\Leftrightarrow4a^2+b^2+c^2\ge2\left(\sqrt{3}-1\right)\left(ab+bc+ca\right)\)

\(\Leftrightarrow4\ge2\left(\sqrt{3}-1\right)\left(ab+bc+ca\right)\)

\(\Leftrightarrow1+\sqrt{3}\ge ab+bc+ca\)

\(\Rightarrow dpcm\)

Bình luận (1)