bat dang thuc

R

rua_it

Cauchy−Schwarz→(∑a2)2=[a2b+c.b+c+b2c+a.c+a+c2a+b.a+b]2≥[∑a4(b+c)](b+c+c+a+a+b)=2[∑a].VT→VT≥∑a32(dpcm)Cauchy-Schwarz \rightarrow (\sum a^2)^2=[\frac{a^2}{\sqrt{b+c}}.\sqrt{b+c}+\frac{b^2}{\sqrt{c+a}}.\sqrt{c+a}+\frac{c^2}{\sqrt{a+b}}.\sqrt{a+b}]^2 \geq [\sum \frac{a^4}{(b+c)}](b+c+c+a+a+b) \\ =2[\sum a].VT \rightarrow VT \geq \frac{\sum a^3}{2}(dpcm)

Tổng quát hóa bài toán:
n∈ Z;a,b,c,x,y>0n \in\ Z; a,b,c,x,y>0 andand n≥1n \geq 1
anbx+cy+bncx+ay+cnax+by≥an−1+bn−1+cn−1x+y\frac{a^n}{bx+cy}+\frac{b^n}{cx+ay}+\frac{c^n}{ax+by} \geq \frac{a^{n-1}+b^{n-1}+c^{n-1}}{x+y}
:D:D:D
 
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