Tìm GTLN-GTNN của tích phân

Vấn đề 13. Tìm GTLN-GTNN của tích phân.
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Lời Giải:
Từ giả thiết $x \cdot f”(x) \geq \mathrm{e}^x+x$ ta có $\displaystyle\int\limits_0^2 x \cdot f”(x)\mathrm{\,d}x \geq \displaystyle\int\limits_0^2 (\mathrm{e}^x+x)\mathrm{\,d}x$.$(1)$.
Đặt $\begin{cases}&u=x\\&\mathrm{d}v=f”(x)\end{cases} \Rightarrow \begin{cases}&\mathrm{d}u=\mathrm{d}x\\&v=f'(x)\end{cases}$.
Khi đó $(1) \Leftrightarrow x \cdot f'(x)\bigg|_0^2 -\displaystyle\int\limits_0^2 f'(x)\mathrm{\,d}x \geq \left(\mathrm{e}^x+\dfrac{x^2}{2}\right)\bigg|_0^2$.
$\begin{aligned}
& \Leftrightarrow x \cdot f'(x)\bigg|_0^2 -f(x)\bigg|_0^2 \geq \left(\mathrm{e}^x+\dfrac{x^2}{2}\right)\bigg|_0^2\\
& \Leftrightarrow \left[2 \cdot f'(2)-0 \cdot f'(0)\right]-\left[f(2)-f(0)\right] \geq \mathrm{e}^2+2-1
\end{aligned}$.
$\Leftrightarrow f(2) \leq 4\mathrm{e}-1$ (do $f'(2)=2\mathrm{e}, f(0)=\mathrm{e}^2$ ).
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Lời Giải:
Từ giả thiết ta có $\dfrac{1}{2} \leq f(x) \leq 2$, suy ra $f(x)+\dfrac{1}{f(x)} \leq \dfrac{5}{2}$.
Suy ra $\displaystyle\int\limits_1^3 \left[f(x)+\dfrac{1}{f(x)}\right]\mathrm{d}x \leq \displaystyle\int\limits_1^3 \dfrac{5}{2}\mathrm{\,d}x \Leftrightarrow \displaystyle\int\limits_1^3 f(x)\mathrm{\,d}x+\displaystyle\int\limits_1^3 \dfrac{1}{f(x)}\mathrm{\,d}x \leq 5 \Leftrightarrow \displaystyle\int\limits_1^3 \dfrac{1}{f(x)}\mathrm{\,d}x \leq 5-\displaystyle\int\limits_1^3 f(x)\mathrm{\,d}x$.
Khi đó $S=\displaystyle\int\limits_1^3 f(x)\mathrm{\,d}x \cdot \displaystyle\int\limits_1^3 \dfrac{1}{f(x)}\mathrm{\,d}x \leq \displaystyle\int\limits_1^3 f(x)\mathrm{\,d}x \cdot \left(5-\displaystyle\int\limits_1^3 f(x)\mathrm{\,d}x\right) \leq \dfrac{25}{4}$.
(dạng $t(5-t)=-t^2+5t=-\left(t-\dfrac{5}{2}\right)^2+\dfrac{25}{4} \leq \dfrac{25}{4}$ )\\
Dấu $”=”$ xảy ra khi và chỉ khi $\displaystyle\int\limits_1^3 f(x)\mathrm{\,d}x=\dfrac{5}{2}$.
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Lời Giải:
Từ giả thiết $f(x)+f'(x) \leq 1$, nhân thêm hai vế cho $\mathrm{e}^x$ để thu được đạo hàm đúng là $\mathrm{e}^xf(x)+\mathrm{e}^xf'(x) \leq \mathrm{e}^x,\,\forall x \in \mathbb{R} \Leftrightarrow \left[\mathrm{e}^xf(x)\right]’ \leq \mathrm{e}^x,\,\forall x \in \mathbb{R}$.
Suy ra $\displaystyle\int\limits_0^1 \left[\mathrm{e}^xf(x)\right]’\mathrm{\,d}x \leq \displaystyle\int\limits_0^1 \mathrm{e}^x\mathrm{\,d}x \Leftrightarrow \left[\mathrm{e}^xf(x)\right]\bigg|_0^1 \leq \mathrm{e}-1 \Leftrightarrow \left[ef(1)-1 \cdot f(0)\right]\bigg|_0^1 \leq \mathrm{e}-1$.
$\xrightarrow{f(0)=0}f(1) \leq \dfrac{\mathrm{e}-1}{\mathrm{e}}$.
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Lời Giải:
Áp dụng bất đẳng thức Cauchy, ta được\\
$M=\displaystyle\int\limits_0^1 \dfrac{1}{\left[f(x)\right]^2}\mathrm{\,d}x+\displaystyle\int\limits_0^1 \left[f'(x)\right]^2\mathrm{\,d}x \geq 2\displaystyle\int\limits_0^1 \dfrac{f'(x)}{f(x)}\mathrm{\,d}x=2\ln f(x)\bigg|\bigg|_0^1 =2\ln \dfrac{f(1)}{f(0)}=2\ln 2018$.
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Lời Giải:
Tích phân từng phần $\displaystyle\int\limits_0^1 (1-x)^2f'(x)\mathrm{\,d}x=-\dfrac{1}{3}$, ta được $f(0)-\dfrac{1}{3}=2\displaystyle\int\limits_0^1 (1-x)f(x)\mathrm{\,d}x$.
Áp dụng bất đẳng thức Cauchy, ta được\\
$2\displaystyle\int\limits_0^1 (1-x)f(x)\mathrm{\,d}x \leq \displaystyle\int\limits_0^1 (1-x)^2\mathrm{\,d}x+\displaystyle\int\limits_0^1 \left[f(x)\right]^2\mathrm{\,d}x$.
Từ đó suy ra $\displaystyle\int\limits_0^1 \left[f(x)\right]^2\mathrm{\,d}x \geq 2\displaystyle\int\limits_0^1 (1-x)f(x)\mathrm{\,d}x-\displaystyle\int\limits_0^1 (1-x)^2\mathrm{\,d}x$.
$\Leftrightarrow \displaystyle\int\limits_0^1 \left[f(x)\right]^2\mathrm{\,d}x \geq f(0)-\dfrac{1}{3}+\dfrac{(1-x)^3}{3}\bigg|_0^1$.
Vậy $\displaystyle\int\limits_0^1 \left[f(x)\right]^2\mathrm{\,d}x-f(0) \geq -\dfrac{2}{3}$.
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Lời Giải:
Với mỗi số thực $\alpha \in \mathbb{R}$ ta có $\left|\displaystyle\int\limits_0^1 \mathrm{e}^xf(x)\mathrm{\,d}x\right|=\left|\displaystyle\int\limits_0^1 \mathrm{e}^xf(x)\mathrm{\,d}x-\displaystyle\int\limits_0^1 \alpha xf(x)\mathrm{\,d}x\right|$.
$=\left|\displaystyle\int\limits_0^1 f(x)\left(\mathrm{e}^x-\alpha x\right)\mathrm{d}x\right| \leq \displaystyle\int\limits_0^1 |f(x)| \cdot \left|\mathrm{e}^x-\alpha x\right|\mathrm{\,d}x \leq \displaystyle\int\limits_0^1 \left|\mathrm{e}^x-\alpha x\right|\mathrm{\,d}x$.
Suy ra $\left|\displaystyle\int\limits_0^1 \mathrm{e}^xf(x)\mathrm{\,d}x\right| \leq \min\limits_{\alpha \in \mathbb{R}} \displaystyle\int\limits_0^1 \left|\mathrm{e}^x-\alpha x\right|\mathrm{\,d}x \leq \min\limits_{\alpha \in [0; 1]} \displaystyle\int\limits_0^1 \left|\mathrm{e}^x-\alpha x\right|\mathrm{\,d}x=\min\limits_{\alpha \in [0; 1]} \left\{\mathrm{e}-1-\dfrac{\alpha}{2}\right\}=\mathrm{e}-\dfrac{3}{2}$.
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Lời Giải:
Từ giả thiết $g(x)=1+\displaystyle\int\limits_0^x f(t)\mathrm{\,d}t,$
ta có $\begin{cases}&g(0)=1\\&g'(x)=f(x)\end{cases}$ và $g(x)>0,\,\forall x \in [0; 1]$.
Theo giả thiết $g(x) \leq \sqrt{f(x)} \to g(x) \leq \sqrt{g'(x)} \Leftrightarrow \dfrac{\sqrt{g'(x)}}{g(x)} \geq 1 \Leftrightarrow \dfrac{g'(x)}{g^2(x)} \geq 1$.
Suy ra $\displaystyle\int\limits_0^t \dfrac{g'(x)}{g^2(x)}\mathrm{\,d}x \geq \displaystyle\int\limits_0^t 1\mathrm{\,d}x,\,\forall t \in [0; 1]\overset{}{\longleftrightarrow}-\dfrac{1}{g(x)}\bigg|_0^t \geq x\bigg|_0^t \Leftrightarrow -\dfrac{1}{g(t)}+\dfrac{1}{g(0)} \geq t \Leftrightarrow \dfrac{1}{g(t)} \leq 1-t$.
Do đó $\displaystyle\int\limits_0^1 \dfrac{1}{g(x)}\mathrm{\,d}x \leq \displaystyle\int\limits_0^1 (1-x)\mathrm{\,d}x=\dfrac{1}{2}$.
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Lời Giải:
Từ giả thiết $g(x)=1+3\displaystyle\int\limits_0^x f(t)\mathrm{\,d}t,$
ta có $\begin{cases}&g(0)=1\\&g'(x)=3f(x)\end{cases}$ và $g(x)>0,\,\forall x \in [0; 1]$.
Theo giả thiết $g(x) \geq f^2(x) \to g(x) \geq \dfrac{\left[g'(x)\right]^2}{9} \Leftrightarrow \dfrac{g'(x)}{2\sqrt{g(x)}} \leq \dfrac{3}{2}$.
Suy ra $\displaystyle\int\limits_0^t \dfrac{g'(x)}{2\sqrt{g(x)}}\mathrm{\,d}x \leq \displaystyle\int\limits_0^t \dfrac{3}{2}\mathrm{\,d}x,\,\forall t \in [0; 1]\overset{}{\longleftrightarrow}\sqrt{g(x)}\bigg|_0^t \leq \dfrac{3}{2}x\bigg|_0^t \Leftrightarrow \sqrt{g(t)}-\sqrt{g(0)} \leq \dfrac{3}{2}t \Leftrightarrow \sqrt{g(t)} \leq \dfrac{3}{2}t+1$.
Do đó $\displaystyle\int\limits_0^1 \sqrt{g(x)}\mathrm{\,d}x \leq \displaystyle\int\limits_0^1 \left(\dfrac{3}{2}x+1\right)\mathrm{d}x=\dfrac{7}{4}$.
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Lời Giải:
Đặt $g(x)=2018+2\displaystyle\int\limits_0^x f(t)\mathrm{\,d}t,$
ta có $\begin{cases}&g(0)=2018\\&g'(x)=2f(x)\end{cases}$ và $g(x)>0,\,\forall x \in [0; 1]$.
Theo giả thiết $g(x) \geq f(x) \to g(x) \geq \dfrac{g'(x)}{2} \Leftrightarrow \dfrac{g'(x)}{g(x)} \leq 2$.
Suy ra $\displaystyle\int\limits_0^t \dfrac{g'(x)}{g(x)}\mathrm{\,d}x \leq \displaystyle\int\limits_0^t 2\mathrm{\,d}x,\,\forall t \in [0; 1]\overset{}{\longleftrightarrow}\ln |g(x)|\bigg|_0^t \leq 2x\bigg|_0^t$.
$\Leftrightarrow \ln g(t)-\ln g(0) \leq 2t \Leftrightarrow \ln g(t) \leq 2t+\ln 2018 \Leftrightarrow g(t) \leq 2018 \cdot \mathrm{e}^{2t}$.
Do đó $\displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x \leq \displaystyle\int\limits_0^1 g(x)\mathrm{\,d}x \leq 2018\displaystyle\int\limits_0^1 \mathrm{e}^{2x}\mathrm{\,d}x=1009\mathrm{e}^{2x}\bigg|_0^1 =1009\mathrm{e}^2-1009$.
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Lời Giải:
Từ giả thiết $g(x)=1+\displaystyle\int\limits_0^{x^2} f(t)\mathrm{\,d}t,$ ta có $\begin{cases}&g(0)=1\\&g'(x)=2xf(x^2)\end{cases}$ và $g(x)>0,\,\forall x \in [0; 1]$.
Theo giả thiết $g(x) \geq 2xf(x^2) \to g(x) \geq g'(x) \Leftrightarrow \dfrac{g'(x)}{g(x)} \leq 1$.
Suy ra $\displaystyle\int\limits_0^t \dfrac{g'(x)}{g(x)}\mathrm{\,d}x \leq \displaystyle\int\limits_0^t 1\mathrm{\,d}x,\,\forall t \in [0; 1]\overset{}{\longleftrightarrow}\ln g(x)\bigg|_0^t \leq x\bigg|_0^t$.
$\Leftrightarrow \ln g(t)-\ln g(0) \leq t \Leftrightarrow \ln g(t) \leq t \Leftrightarrow g(t) \leq \mathrm{e}^t$.
Do đó $\displaystyle\int\limits_0^1 g(x)\mathrm{\,d}x \leq \displaystyle\int\limits_0^1 \mathrm{e}^x\mathrm{\,d}x=\mathrm{e}-1$.
Nhận xét. Gọi $F(t)$ là một nguyên hàm của hàm số $f(t)$ trên đoạn $[0; x^2]$.
Khi đó $g(x)=1+F(t)\bigg|_0^{x^2} =1+F(x^2)-F(0) \to g'(x)=\left[F(x^2)\right]’=(x^2)’F'(x^2)=2xf(x^2)$.
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Lời Giải:
Từ giả thiết $f'(x) \geq f(x)>0,\,\forall x \in [0; 1]$ ta có $\dfrac{f'(x)}{f(x)} \geq 1,\,\forall x \in [0; 1]$.
Suy ra $\displaystyle\int\limits_0^t \dfrac{f'(x)}{f(x)}\mathrm{\,d}x \geq \displaystyle\int\limits_0^t 1\mathrm{\,d}x,\,\forall t \in [0; 1]\overset{}{\longleftrightarrow}\ln f(x)\bigg|_0^t \geq x\bigg|_0^t \Leftrightarrow \ln f(t)-\ln f(0) \geq t \Leftrightarrow f(t) \geq f(0)\mathrm{e}^t$.
Do đó $f(0) \cdot \displaystyle\int\limits_0^1 \dfrac{1}{f(x)}\mathrm{\,d}x \leq \displaystyle\int\limits_0^1 \dfrac{1}{\mathrm{e}^x}\mathrm{\,d}x=\dfrac{\mathrm{e}-1}{\mathrm{e}}$.
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Lời Giải:
Theo Holder\\
$(1)^2=\left[\displaystyle\int\limits_0^{\pi} \cos xf(x)\mathrm{\,d}x\right]^2 \leq \displaystyle\int\limits_0^{\pi} \cos^2x\mathrm{\,d}x \cdot \displaystyle\int\limits_0^{\pi} f^2(x)\mathrm{\,d}x=\dfrac{\pi}{2} \cdot \displaystyle\int\limits_0^{\pi} f^2(x)\mathrm{\,d}x$.
Suy ra $\displaystyle\int\limits_0^{\pi} f^2(x)\mathrm{\,d}x \geq \dfrac{2}{\pi}$. (Đến đây bạn đọc có thể )\\
Dấu $”=”$ xảy ra khi $f(x)=k\cos x$ thay vào $\displaystyle\int\limits_0^{\pi} f(x)\mathrm{\,d}x=1$ ta được\\
$1=\displaystyle\int\limits_0^{\pi} f(x)\mathrm{\,d}x=k\displaystyle\int\limits_0^{\pi} \cos x\mathrm{\,d}x=k \cdot \sin x\bigg|_0^{\pi} =0$.
Điều này hoàn toàn vô lý.\\
Sửa lại cho đúng:\\
Ta có $\displaystyle\int\limits_0^{\pi} f(x)\mathrm{\,d}x=\displaystyle\int\limits_0^{\pi} \cos xf(x)\mathrm{\,d}x=1 \to \begin{cases}&a=\displaystyle\int\limits_0^{\pi} a\cos xf(x)\mathrm{\,d}x\\&b=\displaystyle\int\limits_0^{\pi} bf(x)\mathrm{\,d}x\end{cases}$
với $\begin{cases}&a, b \in \mathbb{R}\\&a^2+b^2>0\end{cases}$.
Theo Holder\\
$(a+b)^2=\left[\displaystyle\int\limits_0^{\pi} \left(a\cos x+b\right)f(x)\mathrm{\,d}x\right]^2 \leq \displaystyle\int\limits_0^{\pi} \left(a\cos x+b\right)^2\mathrm{\,d}x\displaystyle\int\limits_0^{\pi} f^2(x)\mathrm{\,d}x$.
Lại có\\
$\displaystyle\int\limits_0^{\pi} \left(a\cos x+b\right)^2\mathrm{\,d}x=\dfrac{1}{2}\pi (a^2+2b^2)$.
Từ đó suy ra $\displaystyle\int\limits_0^{\pi} f^2(x)\mathrm{\,d}x \geq \dfrac{2(a+b)^2}{\pi (a^2+2b^2)}$ với mọi $a, b \in \mathbb{R}$ và $a^2+b^2>0$.
Do đó $\displaystyle\int\limits_0^{\pi} f^2(x)\mathrm{\,d}x \geq \dfrac{2}{\pi} \cdot \max \left\{\dfrac{(a+b)^2}{a^2+2b^2}\right\}=\dfrac{3}{\pi}$.
Nhận xét: Ta nhân thêm $a, b$ vào giả thiết được gọi là phương pháp biến thiên hằng số.\\
Cách tìm giá trị lớn nhất của $P=\dfrac{(a+b)^2}{a^2+2b^2}$ ta làm như sau:\\
Nếu $b=0 \to P=1$. (chính là đáp án sai mà mình đã làm ở trên)\\
Nếu $b \ne 0 \to P=\dfrac{(a+b)^2}{a^2+2b^2}=\dfrac{\left(\dfrac{a}{b}\right)^2+2\dfrac{a}{b}+1}{\left(\dfrac{a}{b}\right)^2+2} \overset{t=\dfrac{a}{b}}= \dfrac{t^2+2t+1}{t^2+2}$. Tới đây ta khảo sát hàm số hoặc dùng MODE 7 dò tìm. Kết quả thu được GTLN của $P$ bằng $\dfrac{3}{2}$ khi $t=2 \to \dfrac{a}{b}=2 \Leftrightarrow a=2b$.
Vậy dấu $”=”$ để bài toán xảy ra khi $\begin{cases}&a=2b\\&f(x)=b\left(2\cos x+1\right)\end{cases}$
thay ngược lại điều kiện, ta được
$\displaystyle\int\limits_0^{\pi} b\left(2\cos x+1\right)\mathrm{d}x=1 \Leftrightarrow b=\dfrac{1}{\pi} \to f(x)=\dfrac{2\cos x+1}{\pi}$.
Lúc này $\displaystyle\int\limits_0^{\pi} f^2(x)\mathrm{\,d}x=\displaystyle\int\limits_0^{\pi} \left(\dfrac{2\cos x+1}{\pi}\right)\mathrm{d}x=\dfrac{3}{\pi}$.
Cách khác. Đưa về bình phương
Hàm dưới dấu tích phân là $f^2(x), f(x), \cos xf(x)$ nên ta liến kết với $\left[f(x)+\alpha \cos x+\beta \right]^2$.
Với mỗi số thực $\alpha, \beta$ ta có\\
$\displaystyle\int\limits_0^{\pi} \left[f(x)+\alpha \cos x+\beta \right]^2=\displaystyle\int\limits_0^{\pi} f^2(x)\mathrm{\,d}x+2\displaystyle\int\limits_0^{\pi} \left(\alpha \cos x+\beta \right)f(x)\mathrm{\,d}x+\displaystyle\int\limits_0^{\pi} \left(\alpha \cos x+\beta \right)^2\mathrm{\,d}x$.
$=\displaystyle\int\limits_0^{\pi} f^2(x)\mathrm{\,d}x+2\left(\alpha +\beta \right)+\dfrac{\pi}{2}\alpha^2+\pi \beta^2$.
Ta cần tìm $\alpha, \beta$ sao cho $2\left(\alpha +\beta \right)+\dfrac{\pi}{2}\alpha^2+\pi \beta^2$ đạt giá trị nhỏ nhất. Ta có\\
$2\left(\alpha +\beta \right)+\dfrac{\pi}{2}\alpha^2+\pi \beta^2=\dfrac{\pi}{2}\left(\alpha +\dfrac{2}{\pi}\right)^2 + \pi \left(\beta +\dfrac{1}{\pi}\right)^2 – \dfrac{3}{\pi} \geq -\dfrac{3}{\pi}$.
Vậy với $\alpha =-\dfrac{2}{\pi}; \beta =-\dfrac{1}{\pi}$ thì ta có\\
$\displaystyle\int\limits_0^{\pi} \left[f(x)-\dfrac{2}{\pi}\cos x-\dfrac{1}{\pi}\right]^2=\displaystyle\int\limits_0^{\pi} f^2(x)\mathrm{\,d}x-\dfrac{3}{\pi}$.
Suy ra $\displaystyle\int\limits_0^{\pi} f^2(x)\mathrm{\,d}x=\displaystyle\int\limits_0^{\pi} \left[f(x)-\dfrac{2}{\pi}\cos x-\dfrac{1}{\pi}\right]^2+\dfrac{3}{\pi} \geq \dfrac{3}{\pi}$. Dấu $”=”$ xảy ra khi $f(x)=\dfrac{2\cos x+1}{\pi}$.
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Lời Giải:
Liên kết với bình phương $\left[f(x)+\alpha \sin x+\beta \cos x\right]^2$.
Ta có $\displaystyle\int\limits_0^{\pi} \left[f(x)+\alpha \sin x+\beta \cos x\right]^2\mathrm{\,d}x$.
$\begin{aligned}
& =\displaystyle\int\limits_0^{\pi} \left[f(x)\right]^2\mathrm{\,d}x+2\displaystyle\int\limits_0^{\pi} \left(\alpha \sin x+\beta \cos x\right)f(x)\mathrm{\,d}x+\displaystyle\int\limits_0^{\pi} \left(\alpha \sin x+\beta \cos x\right)^2\mathrm{\,d}x\\
& =\displaystyle\int\limits_0^{\pi} \left[f(x)\right]^2\mathrm{\,d}x+2\left(\alpha +\beta \right)+\dfrac{\pi \alpha^2}{2}+\dfrac{\pi \beta^2}{2} \cdot
\end{aligned}$.
Phân tích $2\left(\alpha +\beta \right)+\dfrac{\pi \alpha^2}{2}+\dfrac{\pi \beta^2}{2}=\dfrac{\pi}{2}\left(\alpha +\dfrac{2}{\pi}\right)^2+\dfrac{\pi}{2}\left(\beta +\dfrac{2}{\pi}\right)^2-\dfrac{4}{\pi}$.
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Lời Giải:
Từ giả thiết, ta có $\begin{cases}&a=\displaystyle\int\limits_0^1 a\mathrm{e}^xf(x)\mathrm{\,d}x\\&b=\displaystyle\int\limits_0^1 bf(x)\mathrm{\,d}x\end{cases}$.
Theo Holder\\
$(a+b)^2=\left[\displaystyle\int\limits_0^1 (a\mathrm{e}^x+b)f(x)\mathrm{\,d}x\right]^2 \leq \displaystyle\int\limits_0^1 (a\mathrm{e}^x+b)^2\mathrm{\,d}x\displaystyle\int\limits_0^1 f^2(x)\mathrm{\,d}x$.
Lại có\\
$\displaystyle\int\limits_0^1 (a\mathrm{e}^x+b)^2\mathrm{\,d}x=\displaystyle\int\limits_0^1 \left(a^2\mathrm{e}^{2x}+2ab\mathrm{e}^x+b^2\right)\mathrm{d}x=\dfrac{1}{2}(\mathrm{e}^2-1)a^2+2(\mathrm{e}-1)ab+b^2$.
Suy ra $\displaystyle\int\limits_0^1 f^2(x)\mathrm{\,d}x \geq \dfrac{(a+b)^2}{\dfrac{1}{2}(\mathrm{e}^2-1)a^2+2(\mathrm{e}-1)ab+b^2}$ với mọi $a, b \in \mathbb{R}$ và $a^2+b^2>0$.
Do đó $\displaystyle\int\limits_0^1 f^2(x)\mathrm{\,d}x \geq \max \left\{\dfrac{(a+b)^2}{\dfrac{1}{2}(\mathrm{e}^2-1)a^2+2(\mathrm{e}-1)ab+b^2}\right\}=-1+\dfrac{1}{3-\mathrm{e}}+\dfrac{1}{\mathrm{e}-1} \approx 3, 1316$.
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Lời Giải:
Từ giả thiết, ta có $\begin{cases}&a=\displaystyle\int\limits_0^1 a\sqrt{x}f(x)\mathrm{\,d}x\\&b=\displaystyle\int\limits_0^1 bf(x)\mathrm{\,d}x\end{cases}$.
Theo Holder
$(a+b)^2=\left(\displaystyle\int\limits_0^1 \left(a\sqrt{x}+b\right)f(x)\mathrm{\,d}x\right)^2 \leq \displaystyle\int\limits_0^1 \left(a\sqrt{x}+b\right)^2\mathrm{\,d}x \cdot \displaystyle\int\limits_0^1 f^2(x)\mathrm{\,d}x$.
Lại có\\
$\displaystyle\int\limits_0^1 \left(a\sqrt{x}+b\right)^2\mathrm{\,d}x=\dfrac{a^2}{2}+\dfrac{4ab}{3}+b^2$.
Suy ra $\displaystyle\int\limits_0^1 f^2(x)\mathrm{\,d}x \geq \dfrac{(a+b)^2}{\dfrac{a^2}{2}+\dfrac{4ab}{3}+b^2}$ với mọi $a, b \in \mathbb{R}$ và $a^2+b^2>0$.
Do đó $\displaystyle\int\limits_0^1 f^2(x)\mathrm{\,d}x \geq \max \left\{\dfrac{(a+b)^2}{\dfrac{a^2}{2}+\dfrac{4ab}{3}+b^2}\right\}=3$.
Cách 2. Liên kết với bình phương $\left[f(x)+\alpha \sqrt{x}+\beta \right]^2$.
Ta có $\displaystyle\int\limits_0^{\pi} \left[f(x)+\alpha \sqrt{x}+\beta \right]^2\mathrm{\,d}x$.
$\begin{aligned}
& =\displaystyle\int\limits_0^{\pi} \left[f(x)\right]^2\mathrm{\,d}x+2\displaystyle\int\limits_0^{\pi} \left(\alpha \sqrt{x}+\beta \right)f(x)\mathrm{\,d}x+\displaystyle\int\limits_0^{\pi} \left(\alpha \sqrt{x}+\beta \right)^2\mathrm{\,d}x\\
& =\displaystyle\int\limits_0^{\pi} \left[f(x)\right]^2\mathrm{\,d}x+2\left(\alpha +\beta \right)+\dfrac{\alpha^2}{2}+\dfrac{4}{3}\alpha \beta +\beta^2 \cdot
\end{aligned}$.
Phân tích $2\left(\alpha +\beta \right)+\dfrac{\alpha^2}{2}+\dfrac{4}{3}\alpha \beta +\beta^2=\left(\beta +\dfrac{2}{3}\alpha +1\right)^2+\dfrac{1}{18}\left(\alpha +6\right)^2-3$.
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Lời Giải:
Ta có áp dụng hai lần liên tiếp bất đẳng thức Holder ta được\\
$31^4=\left(\displaystyle\int\limits_1^2 x^3f(x)\mathrm{\,d}x\right)^4=\left(\left[\displaystyle\int\limits_1^2 x^2 \cdot xf(x)\mathrm{\,d}x\right]^2\right)^2 \leq \left(\displaystyle\int\limits_1^2 x^4\mathrm{\,d}x\right)^2\left(\displaystyle\int\limits_1^2 x^2f^2(x)\mathrm{\,d}x\right)^2 \leq \left(\displaystyle\int\limits_1^2 x^4\mathrm{\,d}x\right)^3\displaystyle\int\limits_1^2 f^4(x)\mathrm{\,d}x$.
Suy ra $\displaystyle\int\limits_1^2 f^4(x)\mathrm{\,d}x \geq \dfrac{31^4}{\left(\displaystyle\int\limits_1^2 x^4\mathrm{\,d}x\right)^3}=3875$.
Dấu $”=”$ xảy ra khi $f(x)=kx$ nên $k\displaystyle\int\limits_1^2 x^4\mathrm{\,d}x=31 \Leftrightarrow k=5 \to f(x)=5x^2$.
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Lời Giải:
Ta có$\displaystyle\int\limits_0^1 \left[f”(x)\right]^2\mathrm{\,d}x=3\displaystyle\int\limits_0^1 x^2\mathrm{\,d}x \cdot \displaystyle\int\limits_0^1 \left[f”(x)\right]^2\mathrm{\,d}x \overset{Holder} \geq 3\left(\displaystyle\int\limits_0^1 x \cdot f”(x)\mathrm{\,d}x\right)^2$.
$\underline{\begin{cases}u=x\\ \mathrm{\,d}v=f”(x)\mathrm{\,d}x\end{cases}}= 3\left[f'(1)+f(0)-f(1)\right]^2;$.
$\displaystyle\int\limits_1^2 \left[f”(x)\right]^2\mathrm{\,d}x=3\displaystyle\int\limits_1^2 (x-2)^2\mathrm{\,d}x \cdot \displaystyle\int\limits_1^2 \left[f”(x)\right]^2\mathrm{\,d}x \overset{Holder} \geq 3\left(\displaystyle\int\limits_1^2 (x-2) \cdot f”(x)\mathrm{\,d}x\right)2$.
$\underline{\begin{cases}u=x-2\\ \mathrm{\,d}v=f”(x)\mathrm{\,d}x\end{cases}}= 3\left[-f'(1)+f(2)-f(1)\right]^2$.
Suy ra $\displaystyle\int\limits_0^2 \left[f”(x)\right]^2\mathrm{\,d}x \geq 3\left[f'(1)+f(0)-f(1)\right]^2+3\left[-f'(1)+f(2)-f(1)\right]^2$.
$\geq 3 \cdot \dfrac{\left[f(0)-2f(1)+f(2)\right]^2}{2}=\dfrac{3}{2}$.
Nhận xét: bài giải trên sử dụng bất đẳng thức ở bước cuối là $a^2+b^2 \geq \dfrac{(a+b)^2}{2}$.
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Lời Giải:
Vì $\max \limits_{[1; 3]} |f(x)|=\sqrt{10}\xrightarrow \exists x_0 \in [1; 3]$ sao cho $|f(x_0)|=\sqrt{10}$.
$\xrightarrow{f(1)=0}\exists x_0 \in (1; 3]$ sao cho $|f(x_0)|=\sqrt{10}$.
Theo Holder\\
$\left(\displaystyle\int\limits_1^{x_0} f'(x)\mathrm{\,d}x\right)^2 \leq \displaystyle\int\limits_1^{x_0} 1^2\mathrm{\,d}x \cdot \displaystyle\int\limits_1^{x_0} \left[f'(x)\right]^2\mathrm{\,d}x=(x_0-1) \cdot \displaystyle\int\limits_1^{x_0} \left[f'(x)\right]^2\mathrm{\,d}x$.
Mà $\left(\displaystyle\int\limits_1^{x_0} f'(x)\mathrm{\,d}x\right)^2=\left(f(x)\bigg|_1^{x_0} \right)^2=\left(f(x_0)-f(1)\right)^2=10$.
Từ đó suy ra $\displaystyle\int\limits_1^{x_0} \left[f'(x)\right]^2\mathrm{\,d}x \geq \dfrac{10}{x_0-1}$.
$ \to \displaystyle\int\limits_1^3 \left[f'(x)\right]^2\mathrm{\,d}x \geq \displaystyle\int\limits_1^{x_0} \left[f'(x)\right]^2\mathrm{\,d}x \geq \dfrac{10}{x_0-1} \geq \dfrac{10}{3-1}$.
———————- Hết ———————-