For questions inequalities which involves integrals, like Cauchy-Bunyakovsky-Schwarz or Hölder's inequality. To be used with (inequality) tag.
Questions tagged [integral-inequality]
1106 questions
17
votes
6 answers
How prove this $\int_{a}^{b}f^2(x)dx\le (b-a)^2\int_{a}^{b}[f'(x)]^2dx$
let $f\in C^{(1)}[a,b]$,and such that $f(a)=f(b)=0$, show that
$$\int_{a}^{b}f^2(x)dx\le (b-a)^2\int_{a}^{b}[f'(x)]^2dx\cdots\cdots (1)$$
My try: use Cauchy-Schwarz inequality
we have
$$\int_{a}^{b}[f'(x)]^2dx\int_{a}^{b}x^2dx\ge…
math110
- 93,304
12
votes
4 answers
How prove this $\int_{a}^{b}[f''(x)]^2dx\ge\dfrac{4}{b-a}$
let $f$ on $[a,b]$ two continuously differentiable functions,such
$$f(a)=f(b)=0, f'(a)=1,f'(b)=0,b>a>0$$
show that
$$\int_{a}^{b}[f''(x)]^2dx\ge\dfrac{4}{b-a}$$
My idea: use Cauchy-Schwarz…
math110
- 93,304
11
votes
2 answers
How to prove that $\frac{2}{\pi}\int_{x}^{px}\left(\frac{\sin{t}}{t}\right)^2\,\mathrm dt\le 1-\dfrac{1}{p}$ for $p >1, x\ge0$
Show that for $p>1$ and $x \ge 0$:
$$\dfrac{2}{\pi}\int_{x}^{px}\left(\dfrac{\sin{t}}{t}\right)^2\,\mathrm dt\le 1-\dfrac{1}{p}$$
My idea is to use $$\sin{x}=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\cdots $$
math110
- 93,304
11
votes
5 answers
How prove this inequality $\left(\int_{0}^{1}f(x)dx\right)^2\le\frac{1}{12}\int_{0}^{1}|f'(x)|^2dx$
Let $f\in C^{1}[0,1]$ such that $f(0)=f(1)=0$. Show that
$$\left(\int_{0}^{1}f(x)dx\right)^2\le\dfrac{1}{12}\int_{0}^{1}|f'(x)|^2dx.$$
I think we must use Cauchy-Schwarz inequality
$$\int_{0}^{1}|f'(x)|^2dx\ge…
math110
- 93,304
10
votes
3 answers
How prove this $\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2+\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2\le\frac{\pi}{2}\int_{-\pi}^{+\pi}f^2(x)dx$
Prove or disprove:
if $f(x)\ge 0,\forall x\in [-\pi,\pi]$,show that
$$\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2+\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2\le\dfrac{\pi}{2}\int_{-\pi}^{+\pi}f^2(x)dx$$
I can prove this if $2\pi$ takes the place…
math110
- 93,304
9
votes
1 answer
How prove $e^x|f(x)|\le 2$ if $f(x)=\int_{x}^{x+1}\sin{(e^t)}dt$
let $$f(x)=\int_{x}^{x+1}\sin{(e^t)}dt$$
show that:
$$e^x|f(x)|\le 2$$
My idea: let $$e^t=u$$
then
$$|f(x)|=|\int_{e^x}^{e^{x+1}}\dfrac{1}{u}d\cos{u}|$$
math110
- 93,304
9
votes
0 answers
How to prove Integral inequality with Hardy's inequality
Let $f\in C_{0}^{\infty}((-1,1))$. Prove that for any $t\in (-1,1)$ we have
$$(f(t))^4\le \left(\int_{-1}^{1}\dfrac{[2(1-|x|)f'(x)-f(x)][2(1-|x|)f'(x)+f(x)]}{4(1-|x|^2)}dx\right)\cdot\left(\int_{-1}^{1}|f(x)|^2dx\right)$$
Thank
user246688
8
votes
3 answers
An integral inequality with inverse
Let $f:[0,1]\to [0,1]$ be a non-decreasing concave function, such that $f(0)=0,f(1)=1$. Prove or disprove that :
$$ \int_{0}^{1}(f(x)f^{-1}(x))^2\,\mathrm{d}x\ge \frac{1}{12}$$
A friend posed this to me. He hopes to have solved it, but he is not…
shadow10
- 5,616
8
votes
2 answers
How to prove that $\int_{-\pi}^{+\pi}\cos{(2x)}\cos{(3x)}\cos{(4x)}\cdots\cos{(2005x)}dx$ is positive
show that
$$I=\int_{-\pi}^{+\pi}\cos{(2x)}\cos{(3x)}\cos{(4x)}\cdots\cos{(2005x)}dx>0$$
This problem is my frend ask me,
My try:
$$I=2\int_{0}^{\pi}\cos{(2x)}\cos{(3x)}\cos{(4x)}\cdots\cos{(2005x)}dx$$
and I think maybe use
…
math110
- 93,304
8
votes
1 answer
How prove this integral inequality
let $f:[0,1]\longrightarrow R$ be a differentiable function with continuous derivative such that
$f(1)=0$,show that:
$$4\int_{0}^{1}x^2|f'(x)|^2dx\ge\int_{0}^{1}|f(x)|^2+\left(\int_{0}^{1}|f(x)|dx\right)^2$$
I think it can be use Cauchy-schwarz…
math110
- 93,304
7
votes
1 answer
How prove this integral inequality $6\left(\int_{0}^{1}f(x)dx\right)^2\le 1+ 8\int_{0}^{1}f^3(x)dx$
Let $f$ be a positive-valued,concave function on $[0,1]$,Prove that
$$6\left(\int_{0}^{1}f(x)dx\right)^2\le 1+ 8\int_{0}^{1}f^3(x)dx$$
Let $$A=\int_{0}^{1}f^3(x)dx,B=\left(\int_{0}^{1}f(x)dx\right)^2$$
$$\Longleftrightarrow 6B\le 1+8A$$
let…
math110
- 93,304
7
votes
2 answers
How prove this inequality $ 1-\cos (xy) \le\int_0^xf(t) \sin {(tf(t))}dt + \int_0^y f^{-1}(t) \sin{(tf^{-1}(t))} dt .$
Question:
Let $ f$ be a strictly increasing, continuous function mapping $ I=[0,1]$ onto itself. Prove that the following inequality holds for all pairs $ x,y \in I$: $$ 1-\cos (xy) \le\int_0^xf(t) \sin {(tf(t))}dt + \int_0^y f^{-1}(t)…
math110
- 93,304
7
votes
1 answer
How show that $\dfrac{a^3}{3}\ge\int_{0}^{a}|F(x)-x|^2dx$
Let $F(x)$ be nonnegative and integrable on $[0,a]$ and such that
$$\left(\int_{0}^{t}F(x)dx\right)^2\ge\int_{0}^{t}F^3(x)dx$$
for every $t$ in $[0,a]$,prove or disprove the conjecture:
$$\dfrac{a^3}{3}\ge\int_{0}^{a}|F(x)-x|^2dx$$
This Problem from…
math110
- 93,304
7
votes
3 answers
Given $\int_{\frac13}^{\frac23}f(x)dx=0$, how to prove $4860(\int_0^1f(x)dx)^2\le 11\int_0^1|f''(x)|^2dx$?
Suppose $f\in C^2[0,1]$, and
$\int_{\frac13}^{\frac23}f(x)dx=0$.
Prove that
$$\left(\int_0^1f(x)dx\right)^2\le \frac{11}{4860}\int_0^1|f''(x)|^2dx.$$
This problem is quite similar to Prove the following integral inequality:…
mbfkk
- 1,299
7
votes
1 answer
Prove that $\int_{0}^{1} |p(x)| dx \leq \frac{\pi}{2} $
Let the polynomial $p(x)= a_0 + a_1 x + . . . + a_n x^n$ have coefficients satisfying the relation $$\sum_{i=1}^{n} a_i^{2} = 1$$
Prove that $\int_{0}^{1} |p(x)| dx \leq \frac{\pi}{2} $.
I don't have any idea to prove this inequality, is there any…
user326307
- 315