Inegalitate integrala cu functii concave

[omc]

Fie \( f:[0,1]\to\mathbb{R} \) o functie concava cu \( f(0)=1 \). Sa se arate ca \( \frac32\int_0^1xf(x)dx\leq\int_0^1f(x)dx-\frac14 \). Cand are loc egalitatea?

Dan Marinescu & Viorel Cornea, Lista scurta ONM 2007

[o.m.]

\( xf(t)\geq (f(x)-1)t+x \) for \( 0\leq x\leq 1 \), \( 0\leq t\leq x \).

t is in [0;x]

\( x\int_{0}^{x}f(t)dt\geq (f(x)-1)\int_{0}^{x}tdt+x\int_{0}^{x}dt \)

For x>0 simplify by x and integrate again for x in [0,1]
\( \int_{0}^{1}(\int_{0}^{x}f(t)dt)dx \geq \frac{1}{2}\int_{0}^{1}xf(x)dx+\frac{1}{4}\ (1) \)

LHS is \( \int_{0}^{1}f(t)dt - \int_{0}^{1}tf(t)dt\ (2) \)

(1) and (2) gives the result.

[enescu]

LHS is \( \int_{0}^{1}f(t)dt - \int_{0}^{1}tf(t)dt\ (2) \)

That's not quite obvious.

Altă soluţie:
Sa exploatam concavitatea si faptul ca \( f\left( 0\right)
=1. \) Avem \( f\left( ax+\left( 1-a\right) \cdot0\right) \geq af\left( x\right)+\left( 1-a\right) f\left( 0\right) =af\left( x\right) +1-a, \)
deci
\( \begin{equation}
f\left( ax\right) \geq af\left( x\right) +1-a,(1)
\end{equation} \)
pentru orice \( a,x\in\left[ 0,1\right] . \)

Vom calcula \( \int_{0}^{1}f(x)dx \) facand schimbarea de variabila \( x=t^{\frac{3}{2}}. \)
Rezulta \( dx=\frac{3}{2}t^{\frac{1}{2}}dt \) si
\( \begin{align*}
\int_{0}^{1}f\left( x\right) dx & =\frac{3}{2}\int_{0}^{1}f\left(
t^{\frac{3}{2}}\right) t^{\frac{1}{2}}dt\\
& =\frac{3}{2}\int_{0}^{1}f\left( t^{\frac{1}{2}}\cdot t\right) t^{\frac
{1}{2}}dt\\
& \overset{(1)}{\geq}\frac{3}{2}\int_{0}^{1}\left( t^{\frac{1}{2}}f\left(
t\right) +1-t^{\frac{1}{2}}\right) t^{\frac{1}{2}}dt\\
& =\frac{3}{2}\int_{0}^{1}tf\left( t\right) dt+\frac{3}{2}\int_{0}
^{1}\left( t^{\frac{1}{2}}-t\right) dt.
\end{align*} \)
Cum
\(
\frac{3}{2}\int_{0}^{1}\left( t^{\frac{1}{2}}-t\right) dt=\frac{1}{4},
\)
obtinem inegalitatea dorita.

[Laurentiu Tucaa]

Alta solutie care iese imediat se bazeaza pe o integrare prin parti si apoi un Hermitte-Hadamard(sper ca am scris bine):
Avem \( \int_0^1 xf(x)dx=\int_0^1 xF^{\prime}(x)dx=F(1)-\int_0^1 F(x)dx,F(x)=\int_0^x f(t)dt \).Acum putem aplica HH pt f concava si avem \( F(x)\ge x\cdot\frac{f(x)+f(0)}{2} \).Inegalitatea intre functii se pastreaza prin integrare ,deci\( \int_0^1 xf(x)dx\ge\int_0^1 f(x)dx-\frac{1}{2}\int_0^1 (xf(x)+x)dx \).De aici dupa calcule ajungem la inegalitatea dorita.

[Laurentiu Tucaa]

Peste exact un an de la problema asta ,autorii au generalizat-o .Iata si generalizarea:
Fie \( f:[0,1]\rightarrow\mathbb{R} \),concava ,continua a.i.\( f(0)=1 \).Sa se arate ca :
\( \frac{p+2}{2}\cdot\int_0^1 x^pf(x)dx\le\int_0^1f(x)dx-\frac{p}{2(p+1)},\forall p\ge1 \).
Eu cel putin am facut-o pe aceeasi idee ca in cazul particular p=1 ,insa cred ca se poate rezolva si pe ideea domnului Enescu din rezolvarea aceluiasi caz particular.