[toán 8]Giup to nhe

R

rua_it

[tex]a^3+b^3=(a+b)(a^2+ab+c^2) \geq (a+b).ab[/tex]
[tex]\rightarrow a^3+b^3+abc \geq (a+b+c).ab \geq \frac{abc.(a+b+c)}{c}[/tex]
[tex]\rightarrow \frac{1}{a^3+b^3+abc} \leq \frac{c}{abc.(a+b+c)}[/tex] [tex]\rightarrow \sum_{cyc} \frac{1}{a^3+b^3+abc} \leq \frac{1}{abc.(a+b+c)}.(a+b+c) \leq \frac{1}{abc}(dpcm)[/tex]
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