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a, Ta có : \(\left(a-b\right)^2\ge0< =>a^2-2ab+b^2\ge0< =>a^2+b^2\ge2ab\)
\(\left(a-c\right)^2\ge0< =>a^2-2ac+c^2\ge0< =>a^2+c^2\ge2ac\)
Cộng theo vế hai bất đẳng thức sau : \(a^2+b^2+a^2+c^2\ge2ac+2ab< =>2a^2+b^2+c^2\ge2a\left(b+c\right)\left(đpcm\right)\)
Dấu = xảy ra khi và chỉ khi \(a=b=c\)
a) 2a2 + b2 + c2 ≥ 2a( b + c )
<=> 2a2 + b2 + c2 ≥ 2ab + 2ac
<=> 2a2 + b2 + c2 - 2ab - 2ac ≥ 0
<=> ( a2 - 2ab + b2 ) + ( a2 - 2ac + c2 ) ≥ 0
<=> ( a - b )2 + ( a - c )2 ≥ 0 ( đúng )
Vậy bđt được chứng minh
Đẳng thức xảy ra <=> a = b = c
b) a4 - a3b - ab3 + b4 ≥ 0
<=> a3( a - b ) - b3( a - b ) ≥ 0
<=> ( a - b )( a3 - b3 ) ≥ 0
<=> ( a - b )( a - b )( a2 + ab + b2 ) ≥ 0
<=> ( a - b )2[ ( a2 + ab + 1/4b2 ) + 3/4b2 ] ≥ 0
<=> ( a - b )2[ ( a + 1/2b )2 + 3/4b2 ) ≥ 0 ( đúng )
Vậy bđt được chứng minh
Đẳng thức xảy ra <=> a = b
