9=3(a+b+c) sau đó dùng kỹ thuật tách ghép đối xứng

\(a\sqrt{a+8}+b\sqrt{b+8}+c\sqrt{c+8}\ge9.\)

\("\sqrt{a+8}"\sqrt{b+8}"\sqrt{c+8}"=xyz\Leftrightarrow\left(a,b,c\right)=\left(X^2-8\right)\left(b^2-8\right)\left(c^2-8\right)\) (1)

\(\Leftrightarrow x^2+y^2+z^2=27\) (2)

\(\left(x^2-8\right)x+y\left(y^2-8\right)+z\left(z^2-8\right)\ge9\)

\(x^3+y^3+z^3-8\left(x+y+z\right)\ge9\)

\(\left(x^3+9x\right)+\left(y^3+9y\right)+\left(z^3+9y\right)-17\left(x+y+z\right)\ge6x^2+6y^2+6z^2-17\sqrt{3\left(x^2+y^2+z^2\right)}\)

từ (2) ta có (x^2+y^2+z^2)=27 

\(VT\ge6\left(27\right)-17\sqrt{3\left(27\right)}=162-153=9\)

                                                                                                                         \(\ge\)

\(\Sigma\frac{b+1}{8-\sqrt{a}}\le\Sigma\frac{2\left(b+1\right)}{15-a}=\Sigma\frac{2\left(a+2b+c\right)}{4a+5b+5c}\)(AM-gm)

Đặt \(\left\{\begin{matrix}x=4a+5b+5c\\y=4b+5a+5c\\z=4c+5a+5b\end{matrix}\right.\)suy ra...

Lời giải:

Không mất tính tổng quát, giả sử \(a\geq b\geq c\). Khi đó, hiển nhiên \(\frac{1}{1-\sqrt{a}}\geq \frac{1}{1-\sqrt{b}}\geq\frac{1}{1-\sqrt{c}}\)

Áp dụng BĐT Chebyshev cho hai bộ số trên:

\(3\text{VT}\leq (a+1+b+1+c+1)\left ( \frac{1}{1-\sqrt{a}}+\frac{1}{1-\sqrt{b}}+\frac{1}{1-\sqrt{c}} \right )=6\left ( \frac{1}{1-\sqrt{a}}+\frac{1}{1-\sqrt{b}}+\frac{1}{1-\sqrt{c}} \right )(1)\)

Giờ ta chỉ cần CM \(A= \frac{1}{1-\sqrt{a}}+\frac{1}{1-\sqrt{b}}+\frac{1}{1-\sqrt{c}}\leq \frac{3}{7}\)

Dùng hệ số bất định thôi, ta sẽ CM \(\frac{1}{1-\sqrt{a}}\leq \frac{1}{7}+\frac{1}{98}(a-1)\)

\(\Leftrightarrow (\sqrt{a}-1)^2(6-\sqrt{a})\geq 0\). BĐT này luôn đúng với mọi \(0< a<3\)

Tương tự với các phân thức còn lại \(\Rightarrow \frac{1}{1-\sqrt{a}}+\frac{1}{1-\sqrt{b}}+\frac{1}{1-\sqrt{c}}\leq \frac{3}{7}+\frac{a+b+c-3}{98}=\frac{3}{7}(2)\)

Từ \((1),(2)\Rightarrow \text{VT}\leq \frac{6}{7}\). Dấu $=$ xảy ra khi $a=b=c=1$

Mình đã giải tại đây https://hoc24.vn/hoi-dap/question/169464.html

Theo giả thiết, ta có: \(ab+bc+ca+abc=4\)

\(\Leftrightarrow abc+2\left(ab+bc+ca\right)+4\left(a+b+c\right)+8\)\(=12+\left(ab+bc+ca\right)+4\left(a+b+c\right)\)

\(\Leftrightarrow\left(a+2\right)\left(b+2\right)\left(c+2\right)\)\(=\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(c+2\right)\left(a+2\right)\)

\(\Leftrightarrow\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1\)

\(\Rightarrow a+b+c+6=12\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)-6+a+b+c\)

\(=\left(\frac{12}{a+2}+a-2\right)+\left(\frac{12}{b+2}+b-2\right)+\left(\frac{12}{c+2}+c-2\right)\)

Mặt khác: \(\frac{12}{a+2}+a-2=\frac{12+a^2-4}{a+2}=\frac{a^2+8}{a+2}\)

Tương tự: \(\frac{12}{b+2}+b-2=\frac{b^2+8}{b+2}\); \(\frac{12}{c+2}+c-2=\frac{c^2+8}{c+2}\)

Từ đó suy ra \(a+b+c+6=\frac{a^2+8}{a+2}+\frac{b^2+8}{b+2}+\frac{c^2+8}{c+2}\)

\(\ge\frac{\left(\sqrt{a^2+8}+\sqrt{b^2+8}+\sqrt{c^2+8}\right)^2}{a+b+c+6}\)(Theo BĐT Bunyakovsky dạng phân thức)

\(\Rightarrow\left(a+b+c+6\right)^2\ge\left(\sqrt{a^2+8}+\sqrt{b^2+8}+\sqrt{c^2+8}\right)^2\)

hay \(\sqrt{a^2+8}+\sqrt{b^2+8}+\sqrt{c^2+8}\le a+b+c+6\)

Đẳng thức xảy ra khi a = b = c = 1

Đề sai. Bạn thử với $a=b=c=0,1$ sẽ thấy. 

\(\dfrac{1}{\sqrt{a^3+1}}=\dfrac{1}{\sqrt{\left(a+1\right)\left(a^2-a+1\right)}}\ge\dfrac{2}{a+1+a^2-a+1}=\dfrac{2}{a^2+2}\)

\(\Rightarrow VT\ge\dfrac{2}{a^2+2}+\dfrac{2}{b^2+2}+\dfrac{2}{c^2+2}\)

Do \(abc=8\Rightarrow a^2b^2c^2=64\) , tồn tại các số thực dương x;y;z sao cho:

\(\left(a^2;b^2;c^2\right)=\left(\dfrac{4x}{y};\dfrac{4y}{z};\dfrac{4z}{x}\right)\)

\(\Rightarrow VT\ge\dfrac{2}{\dfrac{4x}{y}+2}+\dfrac{2}{\dfrac{4y}{z}+2}+\dfrac{2}{\dfrac{4z}{x}+2}=\dfrac{y}{2x+y}+\dfrac{z}{2y+z}+\dfrac{x}{2z+x}\)

\(VT\ge\dfrac{x^2}{x^2+2xz}+\dfrac{y^2}{y^2+2xy}+\dfrac{z^2}{z^2+2yz}\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+2xy+2yz+2zx}=1\) (đpcm)

\(VT\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}+\dfrac{b^2}{\sqrt{2\left(a^2+c^2\right)}}+\dfrac{c^2}{\sqrt{2\left(a^2+b^2\right)}}\)

Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\Rightarrow x+y+z=\sqrt{2019}\)

\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{x^2+z^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\) \(\Rightarrow2\sqrt{2}VT\ge\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\)

\(\Rightarrow2\sqrt{2}VT\ge\dfrac{y^2+z^2}{x}+\dfrac{z^2+x^2}{y}+\dfrac{x^2+y^2}{z}-\left(x+y+z\right)\)

\(2\sqrt{2}VT\ge\dfrac{\left(y+z\right)^2}{2x}+\dfrac{\left(z+x\right)^2}{2y}+\dfrac{\left(x+y\right)^2}{2z}-\left(x+y+z\right)\)

\(2\sqrt{2}VT\ge\dfrac{4\left(x+y+z\right)^2}{2x+2y+2z}-\left(x+y+z\right)=x+y+z=\sqrt{2019}\)

\(\Rightarrow VT\ge\dfrac{\sqrt{2019}}{2\sqrt{2}}=\sqrt{\dfrac{2019}{8}}\) (đpcm)

Vì \(ab+bc+ca+abc=4\) nên tồn tại các số x,y,z>0 thỏa mãn \(a=\dfrac{2x}{y+z},b=\dfrac{2y}{x+z},c=\dfrac{2z}{x+y}\).

Viết lại BĐT cần chứng minh:\(\sum\sqrt{\dfrac{4x^2}{\left(y+z\right)^2}+8}\le\sum\dfrac{2x}{y+z}+6\)

\(\Leftrightarrow\sum\dfrac{y+z}{\sqrt{x^2+2\left(y+z\right)^2}+x}\le\dfrac{3}{2}\)(*)

Thật vậy, theo BĐT Bunyakovsky:

\(\left[x^2+2\left(y+z\right)^2\right]\left(1+8\right)\ge\left(x+4y+4z\right)^2\)

\(\Rightarrow VT\le\sum\dfrac{y+z}{\dfrac{x+4y+4z}{3}+x}=\sum\dfrac{3\left(y+z\right)}{4\left(x+y+z\right)}=\dfrac{3}{2}\)

\(\RightarrowĐpcm\)

#proof 2 :[THTT-485]

\(ab+bc+ca+abc=4\Leftrightarrow\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}=1\)

Ta có: \(\dfrac{12}{a+2}+a-2=\dfrac{a^2+8}{a+2}\).Thiết lập tương tự và cộng lại:

\(6+a+b+c=\dfrac{a^2+8}{a+2}+\dfrac{b^2+8}{b+2}+\dfrac{c^2+8}{c+2}\ge\dfrac{\left(\sum\sqrt{a^2+8}\right)^2}{a+b+c+6}\)

\(>>đpcm \)