Question

\(n\ge3;a,b,c>0\)

CMR :

\(\dfrac{1}{a^n\left(b+c\right)}+\dfrac{1}{b^n\left(c+a\right)}+\dfrac{1}{c^n\left(a+b\right)}\ge\dfrac{3}{2}\)

Akai Haruma

Answer 1

Áp dụng BĐT holder cho n bộ 3 số:

\(\left(\sum\dfrac{b^nc^n}{b+c}\right)\left[\sum\left(b+c\right)\right]\left(1+1+1\right)..\left(1+1+1\right)\ge\left(ab+bc+ca\right)^n\)

\(\Leftrightarrow VT\ge\dfrac{\left(ab+bc+ca\right)^n}{3^{n-2}.2.\left(a+b+c\right)}\ge\dfrac{3^{n-2}.3abc\left(a+b+c\right)}{3^{n-2}.2.\left(a+b+c\right)}=\dfrac{3}{2}\)

#Hint:(\(\left\{{}\begin{matrix}ab+bc+ca\ge3\\\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)\end{matrix}\right.\))

BĐT holder thường dùng:

\(\left(a_1^m+a_2^m+...+a_k^m\right)\left(b_1^m+b_2^m+...+b_k^m\right)...\left(c_1^m+...+c_k^m\right)\ge\left(a_1b_1...c_1+a_2.b_2...c_2+...+a_k.b_k...c_k\right)^m\)

trong đó VT có m thừa số từ a đến c

Answer 2

abc = 1 nưa nha