b+ca2+c+ab2+a+bc2 =
b+ca2+c+ab2+a+bc2
<=>
(a+b)(b+c)(c+a)a2(b+c)(c+a)+b2(a+b)(a+c)+c2(a+b)(b+c) =
(a+b)(b+c)(c+a)a2(a+b)(a+c)+b2(b+c)(a+b)+c2(b+c)(a+c)
<=>
(a+b)(b+c)(c+a)a2(c+a)(c−a)+b2(a+b)(a−b)+c2(b+c)(b−c) = 0
Mà (a + b)(b + c)(c + a) khác 0
=> [TEX]a^2(c + a)(c - a) + b^2(a + b)(a - b) + c^2(b + c)(b - c)[/TEX] = 0
<=> [TEX]a^2(c^2 - a^2) + b^2(a^2 - b^2) + c^2(b^2 - c^2)[/TEX] = 0
<=>
a2c2+a2b2+b2c2−a4−b4−c4 = 0
<=> [TEX]a^2c^2 a^2b^2 + b^2c^2[/TEX] = [TEX]a^4 + b^4 + c^4[/TEX]
Sau đó chứng minh BĐT sau:
a4+b4+c4≥a2b2+b2c2+a2c2
Ta có:
a4+b4≥2a2b2
Dấu "=" xảy ra khi a = b
b4+c4≥2b2c2
Dấu "=" xảy ra khi b = c
a4+c4≥2a2c2
Dấu "=" xảy ra khi a = c
=> [TEX]2(a^4 + b^4 + c^4) \geq 2(a^2b^2 + 2b^2c^2 + 2a^2c^2)[/TEX]
<=> [TEX]a^4 + b^4 + c^4 \geq a^2b^2 + b^2c^2 + a^2c^2[/TEX]
Dấu "=" xảy ra <=> a = b = c
Vậy để
b+ca2+c+ab2+a+bc2 =
b+ca2+c+ab2+a+bc2 thì a = b = c