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Chứng minh bất đẳng thức sau: (a^8+b^8+c^8)/3 lớn hơn bằng ( (a+b+c)/3)^8 Cm giúp e với ạ gấp ạ!
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Áp dụng bất đẳng thức $Schwarz$ 3 lần, ta được:
$\dfrac{a^8 + b^8 + c^8}{3} = \dfrac{(a^4)^2}{3} + \dfrac{(b^4)^2}{3} + \dfrac{(c^4)^2}{3}$
$\geq \dfrac{(a^4 + b^4 + c^4)^2}{3 + 3 + 3} = \left(\dfrac{a^4 + b^4 + c^4}{3}\right)^2$
$\geq \left[\left(\dfrac{a^2 + b^2 + c^2}{3}\right)^2\right]^2=\left(\dfrac{a^2 + b^2 + c^2}{3}\right)^4$
$\geq \left[\left(\dfrac{a + b + c}{3}\right)^2\right]^4 =\left(\dfrac{a + b + c}{3}\right)^8$
Dấu = xảy ra $\Leftrightarrow a = b = c$