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[microwavest]

Cho 0<m<n<p; a,b,c>0 sao cho:

 

$\dfrac{a}{m} + \dfrac{b}{n} + \dfrac{c}{p} \geqslant 3$ ; $\dfrac{b}{n} + \dfrac{c}{p} \geqslant 2$ ; $\dfrac{c}{p} \geqslant 1$ 

 Chứng minh $a^2 + b^2 + c^2 \geqslant m^2 + n^2 + p^2$

Edited by hachinh2013, 15-03-2015 - 05:26.

[Lee LOng]

Ta có: $\frac{a}{m}+\frac{b}{n}+\frac{c}{p}\geq 3; \frac{b}{n}+\frac{c}{p}\geq 2; \frac{c}{p}\geq 1\Rightarrow \frac{a^{2}}{m^{2}}+\frac{b^{2}}{n^{2}}+\frac{c^{2}}{p^{2}}\geq 3; \frac{b^{2}}{n^{2}}+\frac{c^{2}}{p^{2}}\geq 2; \frac{c^{2}}{p^{2}}\geq 1$

Áp dụng công thức khai triển Abel có: 

$a^{2}+b^{2}+c^{2}=\frac{c^{2}}{p^{2}}.p^{2}+\frac{b^{2}}{n^{2}}.n^{2}+\frac{a^{2}}{m^{2}}.m^{2}=(p^{2}-n^{2}).\frac{c^{2}}{p^{2}}+(n^{2}-m^{2})(\frac{c^{2}}{p^{2}}+\frac{b^{2}}{n^{2}})+m^{2}(\frac{b^{2}}{n^{2}}+\frac{c^{2}}{p^{2}}+\frac{a^{2}}{m^{2}})\geq m^{2}+n^{2}+p^{2}$