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1a)Xét a2 + 5 - 4a =a2 - 4a + 4+1=(a - 2)2+1\(\ge\)1 hay (a -2)2 + 1 > 0
\(\Rightarrow\)Đpcm
b)Xét 3(a2 + b2 + c2) -(a + b +c)2 =3a2 + 3b2 + 3c2 - a2 - b2 - c2 - 2ab - 2ac - 2bc
=2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc
=(a - b)2 + (a - c)2 + (b - c)2\(\ge\)0 (với mọi a,b,c)
\(\Rightarrow\)Đpcm
2)Xét A=\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(a+c+b\right)=3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\)
áp dụng cô-sy
\(\Rightarrow\)A\(\ge\)9
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=3\)
áp dụng bất đẳng thức AM-GM ta có:
1/a+1/b+1/c>=9/(a+b+c)
=> 1/a+1/b+1/c>=9/1
=> 1/a+1/b+1/c>=9
\(a+b+c>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Leftrightarrow a+b+c>\frac{bc+ac+ab}{abc}\)
\(\Leftrightarrow a+b+c>bc+ac+ab\)
\(\Leftrightarrow a+b+c-bc-ac-ab>0\)
\(\Leftrightarrow abc+a+b+c-bc-ac-ab-abc>0\)
\(\Leftrightarrow abc+a+b+c-bc-ac-ab-1>0\)
\(\Leftrightarrow ab\left(c-1\right)-a\left(c-1\right)-b\left(c-1\right)+\left(c-1\right)>0\)
\(\Leftrightarrow\left(ab-a-b+1\right)\left(c-1\right)>0\)
\(\Rightarrow\left(a-1\right)\left(b-1\right)\left(c-1\right)>0\) (đpcm)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\Leftrightarrow\frac{ab+bc+ac}{abc}=1\Leftrightarrow ab+bc+ac=abc\)
kết hợp gt: a+b+c=1
\(\Rightarrow abc-ab-ac-bc+a+b+c-1=0\Leftrightarrow\left(a-1\right)\left(b-1\right)\left(c-1\right)=0\left(đpcm\right)\)
(a-1)(b-1)(c-1)
=(ab-a-b+1)(c-1)
=abc+a+b+c-ab-bc-ac-1
mà abc=1
=>1+a+b+c-ab-bc-ac-1
=a+b+c-ab-bc-ac
vì abc=1
=>ab=1/c;bc=1/a;ac=1/b
=>(a+b+c)-(1/a+1/b+1/c)
mà a+b+c>1/a+1/b+1/c
=>(a+b+c)-(1/a+1/b+1/c)>0
=>(a-1)(b-1)(c-1)>0
\(\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
\(=\left(ab-a-b+1\right)\left(c-1\right)\)
\(=abc-ac-bc+c-ab+a+b-1\)
\(=-ac-bc+c-ab+a+b\)
Mà abc = 1 nên \(\hept{\begin{cases}ab=\frac{1}{c}\\bc=\frac{1}{a}\\ac=\frac{1}{b}\end{cases}}\)
\(ĐT\Leftrightarrow\left(a+b+c\right)-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)>0\)
(Vì \(\left(a+b+c\right)>\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\))
Vậy \(\left(a-1\right)\left(b-1\right)\left(c-1\right)>0\left(đpcm\right)\)
\(\left(a+b+c\right)^2=1\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=1\)
Ta sẽ chứng minh \(ab+bc+ca=0\)
Thật vậy: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Leftrightarrow ab+bc+ca=0\)
Suy ra \(1=a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2\left(đpcm\right)\)
đúng không ta?