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[Nguyen Huyen Dieu]

cmr V a,b,c,d > 0 ta có bât đẳng thưc :

 

$\frac{a^{6}+b^{6}+c^{6}+d^{6}}{a^{3}+b^{3}+c^{3}+d^{3}}\geqslant \frac{a^{5}+b^{5}+c^{5}+d^{5}}{a^{2}+b^{2}+c^{2}+d^{2}}$

 

dấu "=" xảy ra  <=>  a = b = c = d

Edited by Nguyen Huyen Dieu, 06-08-2021 - 09:08.

[PDF]

cmr V a,b,c,d > 0 ta có bât đẳng thưc :

 

$\frac{a^{6}+b^{6}+c^{6}+d^{6}}{a^{3}+b^{3}+c^{3}+d^{3}}\geqslant \frac{a^{5}+b^{5}+c^{5}+d^{5}}{a^{2}+b^{2}+c^{2}+d^{2}}$

 

dấu "=" xảy ra  <=>  a = b = c = d

Ký hiệu $S(n)=a^{n}+b^{n}+c^{n}+d^{n}$. Khi đó theo BĐT Cauchy-Schwarz ta có $S(n)S(n+2)\geq S(n+1)^{2}$.

Suy ra $$\frac{S(6)}{S(5)}\geq \frac{S(5)}{S(4)}\geq \frac{S(4)}{S(3)}\geq \frac{S(3)}{S(2)},$$

từ đó suy ra đpcm.

Edited by PDF, 06-08-2021 - 15:29.