bất đẳng thức cực khó!!!

A

angleofdarkness

Tham khảo nhé:

Có $a^3>36$ \Rightarrow $a^3-36>0;a>0$ \Rightarrow $\dfrac{a^3-36}{12a}>0$

\Rightarrow $(\dfrac{a}{2}-b-c)^2+\dfrac{a^3-36}{12a}>0$ \Rightarrow $(\dfrac{a}{2}-b-c)^2+\dfrac{a^3-36abc}{12a}>0$ (do abc = 1)

\Rightarrow $(\dfrac{a}{2}-b-c)^2+\dfrac{a^2-36bc}{12}>0$ \Rightarrow $\Big[ \dfrac{a^2}{4}+(b+c)^2-a(b+c) \Big]+\dfrac{a^2}{12}-3bc>0$

\Rightarrow $\dfrac{a^2}{3}+ b^2 + c^2 > ab + bc + ca$
 
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