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Giải:
Biến đổi vế trái, ta được:
(a−1)(b−1)(c−1)(a−1)(b−1)(c−1)
=(ab−a−b+1)(c−1)=(ab−a−b+1)(c−1)
=abc−ab−ac+a−bc+b+c−1=abc−ab−ac+a−bc+b+c−1
=abc−ab−ac−bc+a+b+c−1=abc−ab−ac−bc+a+b+c−1
=abc−(ab+ac+bc)+(a+b+c)−1=abc−(ab+ac+bc)+(a+b+c)−1
Thay ab + ac + bc = abc và a + b + c = 1, ta được:
=abc−abc+1−1=abc−abc+1−1
=0
Bài 2 :
\(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
<=> a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 3ab + 3bc + 3ca
<=> a^2 + b^2 + c^2 = ab + bc + ca
<=> 2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca
<=> ( a - b )^2 + ( b - c )^2 + ( c - a )^2 = 0
<=> a = b = c
1.
\(\Leftrightarrow2a^2+2b^2+18=2ab+6a+6b\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-6a+9\right)+\left(b^2-6b+9\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-3\right)^2+\left(b-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\a-3=0\\b-3=0\end{matrix}\right.\) \(\Leftrightarrow a=b=3\)
2.
\(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c\)
1)
Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)
Dấu "=" xảy ra khi a=b=c
2)
\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)
Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)
\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)
\(a^2+b^2>=2ab\)
\(b^2+c^2>=2bc\)
\(a^2+c^2>=2ac\)
=> \(2\left(a^2+b^2+c^2\right)>=2\left(ab+bc+ac\right)\)DẤU '=' xảy ra khi a=b=c
Vì \(ab+bc+ac=3\) => \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{abc}\)
Đặt \(\frac{1}{a}=x\): \(\frac{1}{b}=y\): \(\frac{1}{c}=z\)=> x+y+z=3xyz
Ta có \(4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+\frac{1}{xyz}\ge13\)
AD BĐT \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\) dấu = khi a=b=c ta có
\(4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{36}{x+y+z}\)=\(\frac{36}{3xyz}=\frac{12}{xyz}\)
=> \(\frac{12}{xyz}+\frac{1}{xyz}\ge13\)
=> \(\frac{13}{xyz}\ge13\)
mà \(3xyz=x+y+z\ge3\sqrt[3]{xyz}\)dấu = khi x=y=z
=> xyz\(\le1\)
=> đpcm
\(\left\{{}\begin{matrix}ab+bc+ca=abc\\a+b+c=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}abc-ab-bc-ca=0\\a+b+c-1=0\end{matrix}\right.\)
\(\left(a-1\right)\left(b-1\right)\left(c-1\right)=\left(a-1\right)\left(bc-b-c+1\right)\)
\(=abc-ab-ac+a-bc+b+c-1\)
\(=\left(abc-ab-bc-ca\right)+\left(a+b+c-1\right)\)
\(=0+0=0\) (ddpcm)
\(VT=\left(a-1\right)\left(b-1\right)\left(c-1\right)\\ =\left(ab-a-b+1\right)\left(c-1\right)\\ =abc-ab-ac+a-bc+b+c-1\\ =abc-\left(ab+bc+ca\right)+\left(a+b+c\right)-1\\ =abc-abc+1-1=0=VP\)