Câu hỏi của TFBoys

Question

Cho các số dương a, b, c thỏa mãn $a^2+b^2+c^2+abc=4.$

CmR: \(\dfrac{a}{\sqrt{\left(b+2\right)\left(c+2\right)}}+\dfrac{b}{\sqrt{\left(c+2\right)\left(a+2\right)}}+\dfrac{c}{\sqrt{\left(a+2\right)\...

Akai Haruma · Accepted Answer

Từng sau nếu tag bạn tag tên dưới câu trả lời nhé, tag thế này không nhận được thông báo đâu .

Bài này tốn sức quá,  đau đầu

Lời giải:

Sử dụng $\sum$ biểu hiện tổng các hoán vị nhé.

Áp dụng BĐT Cauchy-Schwarz:

$\text{VT}=\frac{a^2}{a\sqrt{(b+2)(c+2)}}+\frac{b^2}{b\sqrt{(c+2)(a+2)}}+\frac{c^2}{c\sqrt{(a+2)(b+2)}}\geq \frac{(a+b+c)^2}{\sum a\sqrt{(b+2)(c+2)}}$

Tiếp tục Cauchy-Schwarz:

$(\sum a\sqrt{(b+2)(c+2)})^2\leq (ab+2a+bc+2b+ac+2c)(ac+2a+ba+2b+bc+2c)$

$\Leftrightarrow \sum a\sqrt{(b+2)(c+2)}\leq (ab+bc+ac+2a+2b+2c)$

$\Rightarrow \text{VT}\geq \frac{(a+b+c)^2}{ab+bc+ac+2(a+b+c)}$

Ta sẽ đi chứng minh $\frac{(a+b+c)^2}{ab+bc+ac+2(a+b+c)}\geq 1\Leftrightarrow (a+b+c)^2\geq ab+bc+ac+2(a+b+c)$

$\Leftrightarrow a^2+b^2+c^2+ab+bc+ac\geq 2(a+b+c)$

$\Leftrightarrow (a^2+b^2+c^2)+(a+b+c)^2\geq 4(a+b+c)$

$\Leftrightarrow 4-abc+(a+b+c)^2\geq 4(a+b+c)\Leftrightarrow (a+b+c-2)^2\geq abc$

$\Leftrightarrow a+b+c\geq \sqrt{abc}+2$

Do $a^2+b^2+c^2+abc=4\Rightarrow $

tồn tại $x,y,z>0$ sao cho:$(a,b,c)=\left ( 2\sqrt{\frac{xy}{(z+x)(z+y)}};2\sqrt{\frac{yz}{(x+y)(x+z)}};2\sqrt{\frac{xz}{(y+x)(y+z)}} \right )$

Khi đó , thực hiện vài bước rút gọn, BĐT cần chứng minh chuyển về:

$\sum \sqrt{xy(x+y)}\geq \sqrt{2xyz}+\sqrt{(x+y)(y+z)(x+z)}$

Bình phương hai vế:

$\Leftrightarrow \sum xy(x+y)+2\sqrt{xy^2z(x+y)(y+z)}\geq 2xyz+\prod (x+y)+2\sqrt{2xyz(x+y)(y+z)(x+z)}$

$\Leftrightarrow \sum\sqrt{xy^2z(x+y)(y+z)}\geq 2xyz+\sqrt{2xyz(x+y)(y+z)(x+z)}$

$\Leftrightarrow \sum \sqrt{y(y+x)(y+z)}\geq 2\sqrt{xyz}+\sqrt{2(x+y)(y+z)(x+z)}$ $(\star)$

Đặt biểu thức vế trái là $A$

$A^2=\sum y(y+x)(y+z)+2\sum\sqrt{[y(y+x)(y+z)][x(x+y)(x+z)]}$

$A^2=\sum x^3+\sum xy(x+y)+3xyz+2\sum \sqrt{[(x^2(x+y+z)+xyz][y^2(x+y+z)+xyz]}$

Áp dụng BĐT C-S : $[x^2(x+y+z)+xyz][y^2(x+y+z)+xyz]\geq [xy(x+y+z)+xyz]^2$

$\Rightarrow A^2\geq \sum x^3+\sum xy(x+y)+3xyz+2\sum [xy(x+y+z)+xyz]$

$\Leftrightarrow A^2\geq \sum x^3+3\sum xy(x+y)+15xyz$

Theo BĐT Schur: $\sum x^3+3xyz\geq \sum xy(x+y)$

$\Rightarrow A^2\geq 4\sum xy(x+y)+12xyz=4[\sum xy(x+y)+3xyz]=4(x+y+z)(xy+yz+xz)$

$\Leftrightarrow A\geq 2\sqrt{(x+y+z)(xy+yz+xz)}$

Ta cần chứng minh $2\sqrt{(x+y+z)(xy+yz+xz)}\geq 2\sqrt{xyz}+\sqrt{2(x+y)(y+z)(x+z)}$ (1)

Đặt $\sqrt{(x+y+z)(xy+yz+xz)}=t$, bằng AM-GM dễ thấy $t^2\geq 9xyz$

$(1)\Leftrightarrow 2t\geq 2\sqrt{xyz}+\sqrt{2(t^2-xyz)}$

$\Leftrightarrow 4t^2\geq 4xyz+2(t^2-xyz)+4\sqrt{2xyz(t^2-xyz)}$

$\Leftrightarrow t^2\geq xyz+2\sqrt{2xyz(t^2-xyz)}$ (2)

Áp dụng AM-GM: $2\sqrt{xyz(t^2-xyz)}=\sqrt{8xyz(t^2-xyz)}\leq \frac{8xyz+t^2-xyz}{2}=\frac{7}{2}xyz+\frac{t^2}{2}$

Và $xyz\leq \frac{t^2}{9}$

$\Rightarrow xyz+2\sqrt{2xyz(t^2-xyz)}\leq t^2$

Do đó (2) đúng kéo theo (1) đúng kéo theo (*) đúng nên ta có đpcm.

Dấu bằng xảy ra khi $a=b=c=1$Câu trả lời: Từng sau nếu tag bạn tag tên dưới câu trả lời nhé, tag thế này không nhận được thông báo đâu .