\(A\ge\frac{9}{a+2+b+2+c+2}+\frac{1}{9abc}\)

\(\Rightarrow A\ge\frac{9}{7}+\frac{1}{9abc}\)

Theo BĐT AM-GM ta có: \(1=a+b+c\ge3\sqrt[3]{abc}\)

\(\Rightarrow abc\le\frac{1}{27}\)

\(\Rightarrow\frac{1}{9abc}\ge3\)

Do đó ta có: 

\(A\ge\frac{9}{7}+3=\frac{30}{7}\)

\(A=\text{∑}_{cyc}\frac{a}{a^2+1}+\frac{1}{9abc}=\text{∑}_{cyc}\frac{1}{a+\frac{1}{a}}+\frac{1}{9abc}\)

\(\ge\frac{9}{\text{∑}_{cyc}\left(a+\frac{1}{a}\right)}+\frac{1}{9abc}=P\)

Ta có \(P=\frac{9}{\frac{1}{a+b+c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+\frac{1}{9abc}\)(Vì a + b + c = 1)

\(\ge\frac{9}{\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{9}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+\frac{1}{9abc}\)

\(=\frac{81}{10}.\frac{abc}{ab+bc+ca}+\frac{1}{9abc}\)

\(\Rightarrow P\ge2\sqrt{\frac{3}{ab+bc+ca}}-\frac{21}{10}\ge2\sqrt{\frac{3}{\frac{\left(a+b+c\right)^2}{3}}}-\frac{21}{10}=\frac{39}{10}\)

\(\Rightarrow A\ge P\ge\frac{39}{10}\)

Dấu "=" khi và chỉ khi a = b = c = \(\frac{1}{3}\)

ui..khó qw ~ mún giải lắm nhưng hk đc...e ms lp 7 thoy ak***ahihi^^

nè  đọc cái bất đnagử thức shur và kĩ năng đặt ẩn p-q-r đi là giải ra , nên tìm kiếm trong ộng tổ google đi nhé\

Áp dụng BĐT AM-GM ta có:

\(ab\le\frac{\left(a+b\right)^2}{4}\le\frac{1}{4}\)

Và \(P=a^2+b^2+\frac{1}{a^2}+\frac{1}{b^2}\)

\(=a^2+\frac{1}{16a^2}+b^2+\frac{1}{16b^2}+15\left(\frac{1}{16a^2}+\frac{1}{16b^2}\right)\)

\(\ge2\sqrt{a^2\cdot\frac{1}{16a^2}}+2\sqrt{b^2\cdot\frac{1}{16b^2}}+15\cdot2\sqrt{\frac{1}{16a^2}\cdot\frac{1}{16b^2}}\)

\(=\frac{1}{2}+\frac{1}{2}+15\cdot2\cdot\frac{1}{16ab}\)\(\ge1+15\cdot2\cdot\frac{1}{16\cdot\frac{1}{4}}=\frac{17}{2}\)

Xảy ra khi \(a=b=\frac{1}{2}\)

Mình cũng đang làm 

bài này và cũng chưa

biết cách giải 

mong các bạn giúp với

Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\) \(\left(x,y,z>0\right)\)

Theo đề \(ab+bc+ca=3abc\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{3}{xyz}\)

\(\Rightarrow x+y+z=3\)

Và \(\sqrt{\frac{ab}{a+b+1}}+\sqrt{\frac{bc}{b+c+1}}+\sqrt{\frac{ca}{c+a+1}}\)

\(=\sqrt{\frac{\frac{1}{xy}}{\frac{1}{x}+\frac{1}{y}+1}}+\sqrt{\frac{\frac{1}{yz}}{\frac{1}{y}+\frac{1}{z}+1}}+\sqrt{\frac{\frac{1}{zx}}{\frac{1}{z}+\frac{1}{x}+1}}\)

\(=\frac{1}{\sqrt{x+y+xy}}+\frac{1}{\sqrt{y+z+yz}}+\frac{1}{\sqrt{z+x+zx}}\)

\(\ge\frac{9}{\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}}\) (Cauchy Schwarz)

Ta có: \(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\)

\(=\sqrt{\left(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\right)^2}\)

\(\le\sqrt{3\left(x+y+xy+y+z+yz+z+x+zx\right)}\)

\(=\sqrt{\left[2\left(x+y+z\right)+\left(xy+yz+zx\right)\right]}\)

\(\le\sqrt{6+\frac{\left(x+y+z\right)^2}{3}}=\sqrt{6+\frac{3^2}{3}}=3\)

\(\Rightarrow\sqrt{\frac{ab}{a+b+1}}+\sqrt{\frac{bc}{b+c+1}}+\sqrt{\frac{ca}{c+a+1}}\)

\(\ge\frac{9}{\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}}\ge\frac{9}{3}=3\)

Dấu "=" xảy ra khi: \(x=y=z=1\Rightarrow a=b=c=1\)

cảm ơn bạn :>

Line 11:

...\(=\sqrt{3\left[2\left(x+y+z\right)+xy+yz+zx\right]}\)

\(\le\sqrt{3\left[6+\frac{\left(x+y+z\right)^2}{3}\right]}=\sqrt{3\left(6+3\right)}=3\sqrt{3}\)

\(\Rightarrow VT\ge\frac{9}{3\sqrt{3}}=\sqrt{3}\)

Dấu "=" xảy ra khi: \(a=b=c=1\)

Sửa lại nha\(\frac{19}{b}\)

thay vào \(\frac{1}{a^2+b^2}\)