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

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

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 = \(\frac{1}{3}\)

Ta có:

\(\frac{1}{a+2}+\frac{3}{b+4}\le1-\frac{2}{c+3}\)

\(\Rightarrow1-\frac{1}{a+2}\ge\frac{3}{b+4}+\frac{2}{c+3}\ge2\sqrt{\frac{6}{\left(b+4\right)\left(c+3\right)}}\)

\(\Leftrightarrow\frac{a+1}{a+2}\ge2\sqrt{\frac{6}{\left(b+4\right)\left(c+3\right)}}\left(1\right)\)

Tương tự : \(1-\frac{3}{b+4}\ge\frac{1}{a+2}+\frac{2}{c+3}\ge2\sqrt{\frac{2}{\left(a+2\right)\left(c+3\right)}}\Leftrightarrow\frac{b+1}{b+4}\ge2\sqrt{\frac{2}{\left(a+2\right)\left(c+3\right)}}\left(2\right)\)

và \(\frac{c+1}{c+3}\ge2\sqrt{\frac{3}{\left(a+2\right)\left(b+4\right)}}\left(3\right)\)

Từ 1,2,3  ta có:

\(\frac{a+1}{a+2}.\frac{b+1}{b+4}.\frac{c+1}{c+3}\ge\frac{48}{\left(a+2\right)\left(b+4\right)\left(c+3\right)}\Leftrightarrow Q\ge48\)

Vậy Min Q =48 khi a=1,b=5,c=3

\(\text{⋄}\)Dễ có: \(B\ge\left(3+\frac{4}{a+b}\right)\left(3+\frac{4}{b+c}\right)\left(3+\frac{4}{c+a}\right)\)

\(\text{⋄}\)Đặt \(b+c=x;c+a=y;a+b=z\left(x,y,z>0\right)\)thì \(a=\frac{y+z-x}{2};b=\frac{z+x-y}{2};c=\frac{x+y-z}{2}\)

Giả thiết được viết lại thành: \(x+y+z\le3\)và ta cần tìm giá trị nhỏ nhất của \(\left(3+\frac{4}{x}\right)\left(3+\frac{4}{y}\right)\left(3+\frac{4}{z}\right)\)

\(\text{⋄}\)Ta có: \(\left(3+\frac{4}{x}\right)\left(3+\frac{4}{y}\right)\left(3+\frac{4}{z}\right)=27+36\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+48\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)+\frac{64}{xyz}\)\(\ge27+36.\frac{9}{x+y+z}+48.\frac{27}{\left(x+y+z\right)^2}+64.\frac{27}{\left(x+y+z\right)^3}\ge343\)

Đẳng thức xảy ra khi x = y = z = 1 hay a = b = c = ¹/₂

a, Áp dụng \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)

Áp dụng \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\forall x,y>0\)

Ta có: \(A=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2\ge\frac{\left(2+\frac{1}{a}+\frac{1}{b}\right)^2}{2}\ge\frac{\left(2+\frac{4}{a+b}\right)^2}{2}\ge\frac{\left(2+4\right)^2}{2}=18\)

Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)

b, Áp dụng \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\)

Áp dụng \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\forall x,y,z>0\)

Ta có: \(B=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2+\left(1+\frac{1}{c}\right)^2\ge\frac{\left(3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3}\ge\frac{\left(3+\frac{9}{a+b+c}\right)^2}{3}\ge\frac{\left(3+6\right)^2}{3}=27\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{2}\)

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Sai gì bạn?

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:\(F=\frac{a}{1+b-a}+\frac{b}{1+c-b}+\frac{c}{1+a-c}\)

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

\(=\frac{a^2}{2ab+ac}+\frac{b^2}{2bc+ab}+\frac{c^2}{2ac+bc}\)

\(\ge\frac{\left(a+b+c\right)^2}{2ab+ac+2bc+ab+2ac+bc}=\frac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\)

\(\ge\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\) khi \(a=b=c=\frac{1}{3}\)