Chủ đề này có 2 trả lời

[dark templar]

Bài toán: Cho $a,b,c>0$ thỏa mãn $ab+bc+ca=1$ và 2 số thực $\alpha,\beta \ge 1$.Chứng minh rằng:
$$\sqrt[3]{abc} \le \sqrt[6]{\frac{[1+2(\alpha-1)abc(a+b+c)](a^2+b^2+c^2+2\beta)}{(3+6\alpha)(3+6\beta)}} \le \frac{a+b+c}{3}$$.

[cristianoronaldo]

Bài toán: Cho $a,b,c>0$ thỏa mãn $ab+bc+ca=1$ và 2 số thực $\alpha,\beta \ge 1$.Chứng minh rằng:
$$\sqrt[3]{abc} \le \sqrt[6]{\frac{[1+2(\alpha-1)abc(a+b+c)](a^2+b^2+c^2+2\beta)}{(3+6\alpha)(3+6\beta)}} \le \frac{a+b+c}{3}$$.

Ta có:

$$\sqrt[6]{\frac{[1+2(\alpha-1)abc(a+b+c)](a^2+b^2+c^2+2\beta)}{(3+6\alpha)(3+6\beta)}}=\sqrt[6]{\frac{[\sum a^2b^2+2\alpha abc(a+b+c)](a^2+b^2+c^2+2\beta)}{(3+6\alpha)(3+6\beta)}}$$

$\geqslant \sqrt[6]{\frac{\left ( 1+2\alpha  \right )abc\left ( \sum a \right )\left ( 1+2\beta  \right )\sum ab}{(3+6\alpha)(3+6\beta)}}=\sqrt[6]{\frac{abc\sum a\sum bc}{9}}\geqslant \sqrt[3]{abc}$

Mặt khác, ta có:

$[1+2(\alpha-1)abc(a+b+c)](a^2+b^2+c^2+2\beta)\leqslant \left [ 1+2\left ( \alpha -1 \right )\frac{\left ( \sum ab \right )^2}{3} \right ]\left [ \left ( \sum a \right )^2+2\left ( \beta -1 \right ) \right ]$

$\leqslant \frac{1}{9}\left ( 2\alpha +1 \right )\left ( 2\beta +1 \right )\left ( \sum a \right )^2\leq \frac{1}{81}\left ( 2\alpha +1 \right )\left ( 2\beta +1 \right )\left ( \sum a \right )^6$

$\Rightarrow \sqrt[6]{\frac{[1+2(\alpha-1)abc(a+b+c)](a^2+b^2+c^2+2\beta)}{(3+6\alpha)(3+6\beta)}} \le \frac{a+b+c}{3}$

Đẳng thức xảy ra khi $a=b=c=\sqrt{\dfrac{1}{3}}$.

Bài viết đã được chỉnh sửa nội dung bởi cristianoronaldo: 08-07-2018 - 17:59

[NeverDiex]

Ta có:

6√[1+2(α−1)abc(a+b+c)](a2+b2+c2+2β)(3+6α)(3+6β)=6√[∑a2b2+2αabc(a+b+c)](a2+b2+c2+2β)(3+6α)(3+6β)[1+2(α−1)abc(a+b+c)](a2+b2+c2+2β)(3+6α)(3+6β)6=[∑a2b2+2αabc(a+b+c)](a2+b2+c2+2β)(3+6α)(3+6β)6

⩾6√(1+2α)abc(∑a)(1+2β)∑ab(3+6α)(3+6β)=6√abc∑a∑bc9⩾3√abc⩾(1+2α)abc(∑a)(1+2β)∑ab(3+6α)(3+6β)6=abc∑a∑bc96⩾abc3

Mặt khác, ta có:

[1+2(α−1)abc(a+b+c)](a2+b2+c2+2β)⩽[1+2(α−1)(∑ab)23][(∑a)2+2(β−1)][1+2(α−1)abc(a+b+c)](a2+b2+c2+2β)⩽[1+2(α−1)(∑ab)23][(∑a)2+2(β−1)]

⩽19(2α+1)(2β+1)(∑a)2≤181(2α+1)(2β+1)(∑a)6⩽19(2α+1)(2β+1)(∑a)2≤181(2α+1)(2β+1)(∑a)6

⇒6√[1+2(α−1)abc(a+b+c)](a2+b2+c2+2β)(3+6α)(3+6β)≤a+b+c3⇒[1+2(α−1)abc(a+b+c)](a2+b2+c2+2β)(3+6α)(3+6β)6≤a+b+c3

Đẳng thức xảy ra khi a=b=c=√13a=b=c=13.