Câu hỏi của Quốc Bảo

Question

Cho a , b , c là 3 cạnh của 1 tam giác và $a+b+c=6$

Chứng minh rằng : $3\left(a^2+b^2+c^2\right)+2abc\ge52$

Kuro Kazuya · Accepted Answer

Áp dụng bất đẳng thức tam giác : $\Rightarrow\left\{\begin{matrix}b+c>a\c+a>b\a+b>c\end{matrix}\right.$$\Rightarrow\left\{\begin{matrix}b+c+a>2a\c+a+b>2b\a+b+c>2c\end{matrix}\right.$$\Rightarrow\left\{\begin{matrix}6>2a\6>2b\6>2c\end{matrix}\right.$ $\Rightarrow\left\{\begin{matrix}a< 3\b< 3\c< 3\end{matrix}\right.$ $\Rightarrow\left\{\begin{matrix}3-a>0\3-b>0\3-c>0\end{matrix}\right.$ Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm $\Rightarrow\left(3-a\right)\left(3-b\right)\left(3-c\right)\le\left(\frac{3-a+3-b+3-c}{3}\right)^3$ $\Rightarrow\left(3-a\right)\left(3-b\right)\left(3-c\right)\le\left[\frac{9-\left(a+b+c\right)}{3}\right]^3$ $\Rightarrow\left(3-a\right)\left(3-b\right)\left(3-c\right)\le\left(\frac{9-6}{3}\right)^3$ $\Rightarrow\left(3-a\right)\left(3-b\right)\left(3-c\right)\le1$ $\Rightarrow\left[3\left(3-b\right)-a\left(3-b\right)\right]\left(3-c\right)\le1$ $\Rightarrow\left(9-3b-3a+ab\right)\left(3-c\right)\le1$ $\Rightarrow3\left(9-3b-3a+ab\right)-c\left(9-3b-3a+ab\right)\le1$ $\Rightarrow27-9b-9a+3ab-9c+3bc+3ac-abc\le1$ $\Rightarrow27-9b-9a-9c+3ab+3bc+3ac-abc\le1$ $\Rightarrow27-9\left(a+b+c\right)+3ab+3bc+3ac-abc\le1$ Ta có: $a+b+c=6$ $\Rightarrow-27+3ab+3bc+3ac-abc\le1$ $\Rightarrow-28+3ab+3bc+3ac\le abc$ $\Rightarrow2\left(-28+3ab+3bc+3ac\right)\le2abc$ $\Rightarrow2\left(-28+3ab+3bc+3ac\right)+3\left(a^2+b^2+c^2\right)\le3\left(a^2+b^2+c^2\right)+2abc$ $\Rightarrow-56+6ab+6bc+6ac+3\left(a^2+b^2+c^2\right)\le3\left(a^2+b^2+c^2\right)+2abc$ $\Rightarrow-56+3\left(a^2+b^2+c^2+2ab+2bc+2ac\right)\le3\left(a^2+b^2+c^2\right)+2abc$ $\Rightarrow-56+3\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)+2abc$ Ta có: $a+b+c=6$ $\Rightarrow-56+3.6^2\le3\left(a^2+b^2+c^2\right)+2abc$ $\Rightarrow52\le3\left(a^2+b^2+c^2\right)+2abc$ ( đpcm )Câu trả lời: Áp dụng bất đẳng thức tam giác : $\Rightarrow\left\{\begin{matrix}b+c>a\c+a>b\a+b>c\end{matrix}\right.$$\Rightarrow\left\{\begin{matrix}b+c+a>2a\c+a+b>2b\a+b+c>2c\end{matrix}\right.$$\Rightarrow\left\{\begin{matrix}6>2a\6>2b\6>2c\end{matrix}\right.$ $\Rightarrow\left\{\begin{matrix}a< 3\b< 3\c< 3\end{matrix}\right.$ $\Rightarrow\left\{\begin{matrix}3-a>0\3-b>0\3-c>0\end{matrix}\right.$ Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm $\Rightarrow\left(3-a\right)\left(3-b\right)\left(3-c\right)\le\left(\frac{3-a+3-b+3-c}{3}\right)^3$ $\Rightarrow\left(3-a\right)\left(3-b\right)\left(3-c\right)\le\left[\frac{9-\left(a+b+c\right)}{3}\right]^3$ $\Rightarrow\left(3-a\right)\left(3-b\right)\left(3-c\right)\le\left(\frac{9-6}{3}\right)^3$ $\Rightarrow\left(3-a\right)\left(3-b\right)\left(3-c\right)\le1$ $\Rightarrow\left[3\left(3-b\right)-a\left(3-b\right)\right]\left(3-c\right)\le1$ $\Rightarrow\left(9-3b-3a+ab\right)\left(3-c\right)\le1$ $\Rightarrow3\left(9-3b-3a+ab\right)-c\left(9-3b-3a+ab\right)\le1$ $\Rightarrow27-9b-9a+3ab-9c+3bc+3ac-abc\le1$ $\Rightarrow27-9b-9a-9c+3ab+3bc+3ac-abc\le1$ $\Rightarrow27-9\left(a+b+c\right)+3ab+3bc+3ac-abc\le1$ Ta có: $a+b+c=6$ $\Rightarrow-27+3ab+3bc+3ac-abc\le1$ $\Rightarrow-28+3ab+3bc+3ac\le abc$ $\Rightarrow2\left(-28+3ab+3bc+3ac\right)\le2abc$ $\Rightarrow2\left(-28+3ab+3bc+3ac\right)+3\left(a^2+b^2+c^2\right)\le3\left(a^2+b^2+c^2\right)+2abc$ $\Rightarrow-56+6ab+6bc+6ac+3\left(a^2+b^2+c^2\right)\le3\left(a^2+b^2+c^2\right)+2abc$ $\Rightarrow-56+3\left(a^2+b^2+c^2+2ab+2bc+2ac\right)\le3\left(a^2+b^2+c^2\right)+2abc$ $\Rightarrow-56+3\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)+2abc$ Ta có: $a+b+c=6$ $\Rightarrow-56+3.6^2\le3\left(a^2+b^2+c^2\right)+2abc$ $\Rightarrow52\le3\left(a^2+b^2+c^2\right)+2abc$ ( đpcm )

Akai Haruma · Answer

Cách khác:

Áp dụng BĐT Schur:

$abc\geq (a+b-c)(b+c-a)(c+a-b)=(6-2a)(6-2b)(6-2c)$

$\Rightarrow abc\geq -216+24(ab+bc+ac)-8abc\Leftrightarrow 3abc\geq 8(ab+bc+ac)-72$

Do đó $\text{VT}=3(a^2+b^2+c^2)+2abc\geq 3(a^2+b^2+c^2)+\frac{16}{3}(ab+bc+ac)-48$

$\Leftrightarrow \text{VT}\geq 3(a+b+c)^2-\frac{2}{3}(ab+bc+ac)-48=60-\frac{2}{3}(ab+bc+ac)$

Theo AM-GM $ab+bc+ac\leq \frac{(a+b+c)^2}{3}=12\Rightarrow \text{VT}\geq 52$ (đpcm)

Dấu bằng xảy ra khi $a=b=c=2$Câu trả lời: Cách khác: