Chủ đề này có 2 trả lời

[dark templar]

Bài toán: Cho $a>0;f:[a,+\infty] \to \mathbb{R}$ liên tục thỏa mãn:
$$\int_{a}^{t}f^2(x)dx \le \int_{a}^{t}x^2dx,\forall t \ge a$$
Chứng minh rằng:

• $$\int_{a}^{b}f(x)dx \le \int_{a}^{b}xdx,\forall b \ge a$$
• $$\int_{a}^{b}f(x)dx.\int_{a}^{b}\dfrac{dx}{f(x)} \ge (b-a)^2$$,với $f:[a,b] \to (0;+\infty)$ liên tục.

[Crystal]

Bài toán:
2. $$\int_{a}^{b}f(x)dx.\int_{a}^{b}\dfrac{dx}{f(x)} \ge (b-a)^2$$,với $f:[a,b] \to (0;+\infty)$ liên tục.

Áp dụng BĐT Cauchy - Schwarz, ta có:
$${\left( {b - a} \right)^2} = {\left( {\int\limits_a^b {\sqrt {f\left( x \right)} .\dfrac{1}{{\sqrt {f\left( x \right)} }}dx} } \right)^2} \le \int\limits_a^b {f\left( x \right)} dx\int\limits_a^b {\dfrac{{dx}}{{f\left( x \right)}}} $$
Suy ra đpcm.

[Crystal]

Bài toán: Cho $a>0;f:[a,+\infty] \to \mathbb{R}$ liên tục thỏa mãn:
$$\int_{a}^{t}f^2(x)dx \le \int_{a}^{t}x^2dx,\forall t \ge a$$
Chứng minh rằng:

• $$\int_{a}^{b}f(x)dx \le \int_{a}^{b}xdx,\forall b \ge a$$

Áp dụng BĐT Cauchy - Schwarz, ta có:
$${\left( {\int\limits_a^t {xf\left( x \right)} dx} \right)^2} \leqslant \left( {\int\limits_a^t {{x^2}dx} } \right)\left( {\int\limits_a^t {{f^2}\left( x \right)dx} } \right) \leqslant {\left( {\int\limits_a^t {{x^2}dx} } \right)^2}$$
$$\Rightarrow \int\limits_a^t {xf\left( x \right)} dx \leqslant \int\limits_a^t {{x^2}dx} \Leftrightarrow F\left( t \right) = \int\limits_a^t {{x^2}dx} - \int\limits_a^t {xf\left( x \right)} dx \geqslant 0$$
$$ \Leftrightarrow F\left( t \right) = \int\limits_a^t {x\left( {x - f\left( x \right)} \right)dx \geqslant 0,\,\,\,\forall t \geqslant a} $$
Ta có: $F\left( a \right) = 0;F\left( b \right) \geqslant 0;F'\left( t \right) = t\left( {t - f\left( t \right)} \right)$, khi đó lấy tích phân từng phần:
$$\int\limits_a^b {\left( {x - f\left( x \right)} \right)dx = \int\limits_a^b {\dfrac{1}{x}} } \left( {x\left( {x - f\left( x \right)} \right)} \right) = \int\limits_a^b {\dfrac{1}{x}dF\left( x \right)} = \left. {\dfrac{{F\left( x \right)}}{x}} \right|_a^b + \int\limits_a^b {\dfrac{{F\left( x \right)}}{{{x^2}}}} dx \geqslant 0$$
$$ \Leftrightarrow \int\limits_a^b {\left( {x - f\left( x \right)} \right)dx \geqslant 0 \Leftrightarrow } \int\limits_a^b {xdx \geqslant \int\limits_a^b {f\left( x \right)} } dx,\,\,\forall b \geqslant a$$
Từ đó ta có đpcm.