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\(\text{⋄}\)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
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\(\Leftrightarrow M=\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+â\right)}+\frac{ab}{c^2\left(a+b\right)}\)
áp dụng bđt cauchy ta có:
\(\frac{bc}{a^2\left(b+c\right)}+\frac{b+c}{4bc}\ge\frac{1}{a}\);\(\frac{ca}{b^2\left(c+a\right)}+\frac{c+a}{4ca}\ge\frac{1}{b}\);\(\frac{ab}{c^2\left(a+b\right)}+\frac{a+b}{4ab}\ge\frac{1}{c}\)
\(\Rightarrow M\ge\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\ge3\sqrt[3]{\frac{1}{8abc}}=\frac{3}{2}\)
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Áp dụng BĐT AM-GM ta có :
\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{a+b+c}{abc}\)
\(=\frac{9}{abc\left(a+b+c\right)}\ge\frac{27}{\left(ab+bc+ca\right)^2}\)
Mặt khác theo BĐT AM-GM có :
\(\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)^2\le\left(\frac{a^2+b^2+c^2+2\left(ab+bc+ca\right)^3}{3}\right)=27\)
\(\Rightarrow\frac{27}{\left(ab+bc+ca\right)^2}\ge a^2+b^2+c^2\)
Đặt \(t=a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}=3\)
Xét \(t+\frac{1}{t}=\frac{1}{9}+\frac{1}{t}+\frac{81}{9}.3=\frac{10}{3}\)
Vậy \(MinP=\frac{10}{3}\Leftrightarrow a=b=c=-1\)
Sửa lại chút , vội quá nên đánh lỗi .
Xét \(t+\frac{1}{t}=\frac{1}{9}+\frac{1}{t}+\frac{8t}{9}\ge2\sqrt{\frac{t}{9}.\frac{1}{t}}+\frac{8}{9}.3=\frac{10}{3}\)
\(\Rightarrow MinP=\frac{10}{3}\Leftrightarrow a=b=c=1\)
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đây là cách của t. t nghĩ nó đơn giản hơn lời giải đó
Ta có : \(\left(a+b\right)^4\le\left(a+b\right)^4+\left(a-b\right)^4=2a^4+2b^4+12a^2b^2\)
\(=2a^4+2b^4+\frac{32}{3}a^2b^2+\frac{2}{3}.2a^2b^2\le2a^4+2b^4+\frac{32}{3}a^2b^2+\frac{2}{3}\left(a^4+b^4\right)\)( Cô-si )
\(=\frac{8}{3}a^4+\frac{8}{3}b^4+\frac{32}{3}a^2b^2\)
Tương tự : \(\left(b+c\right)^4\le\frac{8}{3}b^4+\frac{8}{3}c^4+\frac{32}{3}b^2c^2\); \(\left(a+c\right)^4\le\frac{8}{3}a^4+\frac{8}{3}c^4+\frac{32}{3}a^2c^2\)
Áp dụng BĐT Cô-si dạng Engel, ta có :
\(\left(\frac{a}{a+b}\right)^4+\left(\frac{b}{b+c}\right)^4+\left(\frac{c}{c+a}\right)^4\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a+b\right)^4+\left(b+c\right)^4+\left(c+a\right)^4}\)
\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{\frac{16}{3}\left(a^4+b^4+c^4\right)+\frac{32}{3}\left(a^2b^2+b^2c^2+a^2c^2\right)}=\frac{\left(a^2+b^2+c^2\right)^2}{\frac{16}{3}\left(a^2+b^2+c^2\right)^2}=\frac{3}{16}\)
Dấu "=" xảy ra \(\Leftrightarrow\)x = y = z
Vậy GTNN của P là \(\frac{3}{16}\)\(\Leftrightarrow\)x = y = z
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ta có:
\(M+4=\left(\frac{a-d}{d+b}+1\right)+\left(\frac{d-b}{b+c}+1\right)+\left(\frac{b-c}{c+a}+1\right)+\left(\frac{c-a}{d+a}+1\right)\)
\(=\frac{a+b}{b+d}+\frac{c+d}{b+c}+\frac{a+b}{c+a}+\frac{c+d}{d+a}\)
\(=\left(a+b\right)\left(\frac{1}{b+d}+\frac{1}{c+a}\right)+\left(c+d\right)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)\ge\left(a+b\right).\frac{4}{a+b+c+d}+\left(c+d\right).\frac{4}{a+b+c+d}\)
\(=\frac{4\left(a+b+c+d\right)}{a+b+c+d}=4\)
\(\Rightarrow M+4\ge4\Rightarrow M\ge0\)
vậy min M=0 khi a=b=c=d
\(P=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(=\frac{a^2}{ab+ac}+\frac{b^2}{bc+ba}+\frac{c^2}{ac+bc}\)
\(\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
\(\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)
dấu "=" xảy ra tại a=b=c
Cách 2
\(P+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge\left(a+b+c\right)\cdot\frac{9}{2\left(a+b+c\right)}=\frac{9}{2}\)
\(\Rightarrow P\ge\frac{3}{2}\Leftrightarrow a=b=c\)
\(P=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\left(1\right)\)
Đặt \(\hept{\begin{cases}b+c=x\\c+a=y\\a+b=z\end{cases}\left(x,y,z>0\right)}\)
\(\Rightarrow a=\frac{y+z-x}{2}\);\(b=\frac{z+x-y}{2}\);\(c=\frac{x+y-z}{2}\)
\(\left(1\right)\)trở thành \(\frac{y+z-x}{2x}+\frac{z+x-y}{2y}+\frac{x+y-z}{2z}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{y}{2x}+\frac{z}{2x}-\frac{1}{2}+\frac{z}{2y}+\frac{x}{2y}-\frac{1}{2}+\frac{x}{2z}+\frac{y}{2z}-\frac{1}{2}\ge\frac{3}{2}\)
\(\Leftrightarrow\left(\frac{y}{2x}+\frac{x}{2y}\right)+\left(\frac{z}{2x}+\frac{x}{2z}\right)+\left(\frac{z}{2y}+\frac{y}{2z}\right)\ge3\)
Vì \(\frac{y}{2x}+\frac{x}{2y}\ge2\sqrt{\frac{y}{2x}.\frac{x}{2y}}=1\)( bđt AM-GM)
CMTT \(\frac{z}{2x}+\frac{x}{2z}\ge1\)và \(\frac{z}{2y}+\frac{y}{2z}\ge1\)
rồi cộng vào là xong
Dấu"="xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{y}{2x}=\frac{x}{2y}\\\frac{z}{2x}=\frac{x}{2z}\\\frac{z}{2y}=\frac{y}{2z}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}2x^2=2y^2\\2z^2=2x^2\\2y^2=2z^2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=y\\z=x\\y=z\end{cases}\Leftrightarrow}x=y=z\)
Vậy \(P_{min}=\frac{3}{2}\Leftrightarrow x=y=z\)