bất đẳng thức khó

T

truongduong9083

Ta có
[TEX](ab+bc+ca)^2= (\sqrt{a}.\sqrt{a}b+\sqrt{b}.\sqrt{b}c+ \sqrt{c}. \sqrt{c}a)^2\leq (a+b+c)(ab^2+bc^2+ca^2) \leq \sqrt{3(a^2+b^2+c^2)}(ab^2+bc^2+ca^2)[/TEX]
[TEX]\Rightarrow (ab+bc+ca)^2 \leq 3(ab^2+bc^2+ca^2)[/TEX]
Suy ra
[TEX]P \geq \frac{1}{3} [/TEX]
Dấu bằng xảy ra khi a = b= c = 1
 
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