3 replies to this topic

[skykute]

$\left ( \frac{12}{5} \right )^{x}+\left ( \frac{15}{4} \right )^{x}+\left ( \frac{20}{3} \right )^{x}\geq 3^{x}+4^{x}+5^{x}$

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[Minhnguyenthe333]

$\left ( \frac{12}{5} \right )^{x}+\left ( \frac{15}{4} \right )^{x}+\left ( \frac{20}{3} \right )^{x}\geq 3^{x}+4^{x}+5^{x}$

Dùng bđt Cauchy ta có:
$( \frac{12}{5})^x+(\frac{15}{4})^x\geqslant 2\sqrt{(\frac{12}{5})^x.(\frac{15}{4})^x}=2.3^x$
Cộng vế theo vế các bđt còn lại thì ta có:
$2[\left ( \frac{12}{5} \right )^{x}+\left ( \frac{15}{4} \right )^{x}+\left ( \frac{20}{3} \right )^{x}]\geqslant 2(3^{x}+4^{x}+5^{x})$
hay $\left ( \frac{12}{5} \right )^{x}+\left ( \frac{15}{4} \right )^{x}+\left ( \frac{20}{3} \right )^{x}\geqslant 3^{x}+4^{x}+5^{x}$ (đpcm)

Edited by Minhnguyenthe333, 05-02-2016 - 22:15.

[quangnghia]

$\left ( \frac{12}{5} \right )^{x}+\left ( \frac{15}{4} \right )^{x}+\left ( \frac{20}{3} \right )^{x}\geq 3^{x}+4^{x}+5^{x}$

Cái bài bất đẳng thức bé bé xinh xinh bạn ghi bên dưới thì mình làm nó như sau:

Ta có $\sqrt{\frac{a^{3}}{b^{3}}}+\sqrt{\frac{a^{3}}{b^{3}}}+1\geq 3\frac{a}{b}$

$\sqrt{\frac{b^{3}}{c^{3}}}+\sqrt{\frac{b^{3}}{c^{3}}}+1\geq 3\frac{b}{c}$

$\sqrt{\frac{c^{3}}{a^{3}}}+\sqrt{\frac{c^{3}}{a^{3}}}+1\geq 3\frac{c}{a}$

Cộng theo vế ta có

$2(\sqrt{\frac{a^{3}}{b^{3}}}+\sqrt{\frac{b^{3}}{c^{3}}}+\sqrt{\frac{c^{3}}{a^{3}}})+3\geq 3(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\geq 2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})+3$

Vậy $\sqrt{\frac{a^{3}}{b^{3}}}+\sqrt{\frac{b^{3}}{c^{3}}}+\sqrt{\frac{c^{3}}{a^{3}}}\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$

[I Love MC]

Mở rộng 
Cho $x=\frac{a}{b},y=\frac{b}{c},z=\frac{c}{a},a,b,c$ đều dương 
Khi đó với mọi $n \in \mathbb{N}$ thì  
$x^{n+1}+y^{n+1}+z^{n+1} \ge x^n+y^n+z^n$