[ Toán 8 ] chứng minh BĐT

S

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a) Theo bất đẳng thức tam giác ta có:

$a < b+c \rightarrow 2a <a+b+c=2 \rightarrow a<1$

Tương tự $b<1;c<1$

b) Vì $a< 1\rightarrow a-1<0 \\ b<1 \rightarrow a-1 < 0 \\ c<1 \rightarrow
c-1 <0$

$\leftrightarrow (a-1)(b-1)(c-1) <0 \leftrightarrow abc -(ab+bc+ca)+
(a+b+c)-1 <0$

$\leftrightarrow abc <ab+bc+ca-(a+b+c)+1=ab+bc+ca-1$

$\leftrightarrow 2abc < 2ab+2bc+2ca-2$

$\leftrightarrow 2abc +a^2+b^2+c^2 <(a+b+c)^2 -2$

$\leftrightarrow 2abc+a^2+b^2+c^2 < 4-2=2(dpcm)$
 
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