Đề bài:
Lời giải:
Đặt: $\displaystyle \cos\frac{\pi}{7}=\alpha$.
Áp dụng Bất đẳng thức AM-GM ta có:
$4\alpha^2a^2+b^2\ge4\alpha ab$
$\displaystyle (4\alpha^2-1)b^2+\frac{4\alpha^2}{4\alpha^2-1}c^2\ge4\alpha bc$
$\displaystyle \frac{16\alpha^4-8\alpha^2}{4\alpha^2-1}(c^2+d^2)\ge\frac{32\alpha^4-16\alpha^2}{4\alpha^2-1}cd$
$\displaystyle \frac{4\alpha^2}{4\alpha^2-1}d^2+(4\alpha^2-1)e^2\ge4\alpha de$
$e^2+4\alpha^2f^2\ge4\alpha ef$
Dễ dàng chứng minh được: $\displaystyle \frac{32\alpha^4-16\alpha^2}{4\alpha^2-1}=4\alpha$
Suy ra: $4\alpha^2(a^2+b^2+c^2+d^2+e^2+f^2)\ge4\alpha(ab+bc+cd+de+ef)\ge4\alpha$
$\displaystyle \Rightarrow a^2+b^2+c^2+d^2+e^2+f^2\ge\frac{1}{\alpha}=\frac{1}{\displaystyle \cos\frac{\pi}{7}}$
Cho $a,b,c,d,e,f \in R$ sao cho $ab+bc+cd+de+ef\geq 1$. CMR:
$\displaystyle a^2+b^2+c^2+d^2+e^2+f^2 \geq \frac{1}{\cos \frac{\pi }{7}}$
Đặt: $\displaystyle \cos\frac{\pi}{7}=\alpha$.
Áp dụng Bất đẳng thức AM-GM ta có:
$4\alpha^2a^2+b^2\ge4\alpha ab$
$\displaystyle (4\alpha^2-1)b^2+\frac{4\alpha^2}{4\alpha^2-1}c^2\ge4\alpha bc$
$\displaystyle \frac{16\alpha^4-8\alpha^2}{4\alpha^2-1}(c^2+d^2)\ge\frac{32\alpha^4-16\alpha^2}{4\alpha^2-1}cd$
$\displaystyle \frac{4\alpha^2}{4\alpha^2-1}d^2+(4\alpha^2-1)e^2\ge4\alpha de$
$e^2+4\alpha^2f^2\ge4\alpha ef$
Dễ dàng chứng minh được: $\displaystyle \frac{32\alpha^4-16\alpha^2}{4\alpha^2-1}=4\alpha$
Suy ra: $4\alpha^2(a^2+b^2+c^2+d^2+e^2+f^2)\ge4\alpha(ab+bc+cd+de+ef)\ge4\alpha$
$\displaystyle \Rightarrow a^2+b^2+c^2+d^2+e^2+f^2\ge\frac{1}{\alpha}=\frac{1}{\displaystyle \cos\frac{\pi}{7}}$
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