Bài 104559

Question

Cho $\begin{cases}a_{1},a_{2},...,a_{n},b_{1},b_{2},...,b_{n}>0 \\ \frac{1}{p}+\frac{1}{q}=1 \\ n \in Z,n \geq 2; p,q>0\end{cases}$
Chứng minh rằng:
$a_{1}b_{1}+a_{2}b_{2}+...+a_{n}b_{n}\leq \sqrt [p]{a^{p}_{1}+a^{p}_{2}+...+a^{p}_{n}}.\sqrt [q]{b^{q}_{1}+b^{q}_{2}+...+b^{q}_{n}}$

score 0 · Answer 1

vì:$\frac{1}{p}+\frac{1}{q}=1 $
$\Rightarrow \frac{1}{p},\frac{1}{q}<1$
$\Rightarrow p,q>1$
Xét $f(x)=x^{p}(x>0)$
$f'(x)=p.x^{p-1}$
$f''(x)=p.(p-1)x^{p-2}>0$
Chọn: $\begin{cases}\alpha_{i}=b_{i}^{q}>0 \\ x_{i}=a_{i}b_{i}^{1-q}>0\end{cases}(\forall i=1,2,...,n)$
Theo BĐT JenSen:
$\frac{\alpha_{1}f(x_{1})+\alpha_{2}f(x_{2})+...+\alpha_{n}f(x_{n})}{\alpha_{1}+\alpha_{2}+...+\alpha_{n}}\geq f(\frac{\alpha_{1}x_{1}+\alpha_{2}x_{2}+...+\alpha_{n}x_{n}}{\alpha_{1}+\alpha_{2}+...+\alpha_{n}})$
$\Rightarrow \frac{b^{q}_{1}.a^{p}_{1}b^{(1-q)p}_{1}+b^{q}_{2}.a^{p}_{2}b^{(1-q)p}_{2}+...+b^{q}_{n}.a^{p}_{n}b^{(1-q)p}_{n}}{b^{q}_{1}+b^{q}_{2}+...+b^{q}_{n}} \geq (\frac{a_{1}b_{1}+a_{2}b_{2}+...+a_{n}b_{n}}{b^{q}_{1}+b^{q}_{2}+...+b^{q}_{n}})^{p}$
$\Rightarrow (\frac{a_{1}b_{1}+a_{2}b_{2}+...+a_{n}b_{n}}{b^{q}_{1}+b^{q}_{2}+...+b^{q}_{n}})^{p} \leq (b^{q}_{1}+b^{q}_{2}+...+b^{q}_{n})^{p-1}.(a^{p}_{1}+a^{p}_{2}+...+a^{p}_{n})$ (vì: $q+(1-q)p=0$)
$\Rightarrow a_{1}b_{1}+a_{2}b_{2}+...+a_{n}b_{n}\leq \sqrt [p]{a^{p}_{1}+a^{p}_{2}+...+a^{p}_{n}}.\sqrt [q]{b^{q}_{1}+b^{q}_{2}+...+b^{q}_{n}}$
$\Rightarrow $ (ĐPCM)

Bài 104559

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Bài 104559

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