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\(P=\dfrac{bc}{\dfrac{a^2bc}{c}+\dfrac{a^2bc}{b}}+\dfrac{ca}{\dfrac{b^2ac}{a}+\dfrac{b^2ac}{c}}+\dfrac{ab}{\dfrac{c^2ab}{b}+\dfrac{c^2ab}{a}}=\dfrac{\left(bc\right)^2}{a^2b^2c+a^2bc^2}+\dfrac{\left(ca\right)^2}{b^2a^2c+b^2ac^2}+\dfrac{\left(ab\right)^2}{c^2a^2b+c^2ab^2}=\dfrac{\left(bc\right)^2}{ab+ac}+\dfrac{\left(ca\right)^2}{ba+bc}+\dfrac{\left(ab\right)^2}{ca+cb}\ge\dfrac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}=\dfrac{ab+bc+ca}{2}\ge\dfrac{3\sqrt[3]{\left(abc\right)^2}}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra <=> a = b = c = 1
Lời giải:
\(a+b+c+ab+bc+ac+abc=0\)
\(\Leftrightarrow (a+b+ab+1)+c+bc+ac+abc=1\)
\(\Leftrightarrow (a+b+ab+1)+c(1+b+a+ab)=1\)
\(\Leftrightarrow (a+1)(b+1)+c(a+1)(b+1)=1\)
\(\Leftrightarrow (a+1)(b+1)(c+1)=1\)
Đặt \((a+1,b+1,c+1)=(x,y,z)\Rightarrow (a,b,c)=(x-1,y-1,z-1)\) và \(xyz=1\)
Khi đó:
\(P=\frac{1}{3+2(x-1)+y-1+(x-1)(y-1)}+\frac{1}{3+2(y-1)+z-1+(y-1)(z-1)}+\frac{1}{3+2(z-1)+x-1+(x-1)(z-1)}\)
\(=\frac{1}{x+xy+1}+\frac{1}{y+yz+1}+\frac{1}{z+xz+1}\)
\(=\frac{yz}{xyz+xy.yz+yz}+\frac{1}{y+yz+1}+\frac{y}{zy+xz.y+y}\)
\(=\frac{yz}{1+y+yz}+\frac{1}{y+yz+1}+\frac{y}{yz+1+y}=\frac{yz+1+y}{yz+1+y}=1\)
Ta có đpcm.
a)Bunhia:
\(\left(1+2\right)\left(b^2+2a^2\right)\ge\left(1.b+\sqrt{2}.\sqrt{2}a\right)^2=\left(b+2a\right)^2\)
b)\(ab+bc+ca=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng bđt câu a
=>VT\(\ge\)\(\dfrac{b+2a}{\sqrt{3}ab}+\dfrac{c+2b}{\sqrt{3}bc}+\dfrac{a+2c}{\sqrt{3}ca}\)
\(\Leftrightarrow VT\ge\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{1}{c}+\dfrac{2}{a}=3=VP\)
Tự tìm dấu "="
Nguyễn Việt LâmMashiro ShiinaBNguyễn Thanh HằngonkingCẩm MịcFa CTRẦN MINH HOÀNGhâu DehQuân Tạ MinhTrương Thị Hải Anh
Áp dụng BĐt cô-si, ta có \(\frac{2\left(a+b\right)^2}{2a+3b}\ge\frac{8ab}{2a+3b}=\frac{8}{\frac{2}{b}+\frac{3}{a}}\)
\(\frac{\left(b+2c\right)^2}{2b+c}\ge\frac{8bc}{2b+c}=\frac{8}{\frac{2}{c}+\frac{1}{b}}\)
\(\frac{\left(2c+a\right)^2}{c+2a}\ge\frac{8ac}{c+2a}\ge\frac{8}{\frac{1}{a}+\frac{2}{c}}\)
Cộng 3 cái vào, ta có
A\(\ge8\left(\frac{1}{\frac{2}{b}+\frac{3}{a}}+\frac{1}{\frac{1}{b}+\frac{2}{c}}+\frac{1}{\frac{1}{a}+\frac{2}{c}}\right)\ge8\left(\frac{9}{\frac{3}{b}+\frac{4}{c}+\frac{4}{a}}\right)=8.\frac{9}{3}=24\)
Vậy A min = 24
Neetkun ^^