Câu hỏi của Quách Nguyễn Sông Trà

Question

Cho ba so thuc khong am a,b,c thoa man a+b+c=3. Chung minh a^3+b^3+c^3+ab+ac+bc>=6

Nguyễn Việt Lâm · Accepted Answer

$a^3+a^3+1\ge3a^2\Rightarrow a^3+\frac{1}{2}\ge\frac{3}{2}a^2$

$\Rightarrow VT+\frac{3}{2}\ge\frac{3}{2}a^2+\frac{3}{2}b^2+\frac{3}{2}c^2+ab+bc+ca$

$\Rightarrow VT+\frac{3}{2}\ge a^2+b^2+c^2+\frac{1}{2}\left(a+b+c\right)^2$

$\Rightarrow VT+\frac{3}{2}\ge\frac{1}{3}\left(a+b+c\right)^2+\frac{1}{2}\left(a+b+c\right)^2=\frac{15}{2}$

$\Rightarrow VT\ge\frac{15}{2}-\frac{3}{2}=6$

Dấu "=" xảy ra khi $a=b=c=1$Câu trả lời: $a^3+a^3+1\ge3a^2\Rightarrow a^3+\frac{1}{2}\ge\frac{3}{2}a^2$

$\Rightarrow VT+\frac{3}{2}\ge\frac{3}{2}a^2+\frac{3}{2}b^2+\frac{3}{2}c^2+ab+bc+ca$

$\Rightarrow VT+\frac{3}{2}\ge a^2+b^2+c^2+\frac{1}{2}\left(a+b+c\right)^2$

$\Rightarrow VT+\frac{3}{2}\ge\frac{1}{3}\left(a+b+c\right)^2+\frac{1}{2}\left(a+b+c\right)^2=\frac{15}{2}$

$\Rightarrow VT\ge\frac{15}{2}-\frac{3}{2}=6$

Dấu "=" xảy ra khi $a=b=c=1$

tth_new · Answer

Sau khi đưa BĐT về dạng thuần nhất ta có:

$VT-VP=\frac{1}{18} \sum\limits_{cyc} (7a+7b+c)(a-b)^2  \geq 0$Câu trả lời: Sau khi đưa BĐT về dạng thuần nhất ta có:

$VT-VP=\frac{1}{18} \sum\limits_{cyc} (7a+7b+c)(a-b)^2  \geq 0$

Violympic toán 9